CBSE Class 10 Mathematics Some Applications of Trigonometry Worksheet Set 09

1. If cotΘ = 15/8, evaluate (2 + 2sinΘ)(1 – sinΘ)

(1 + cosΘ)(2 – 2cosΘ) (225/64)

2. If tan A = 2 . Evaluate secA sinA + tan2 A – cosec A

3. In a ΔABC, right angled at A,if tan C = √3, find the value of sinB cosC + cosB sinC (1)

4. in ΔPQR, right angled at Q, QR = 6 cm, <QPR = 60˚. Find the length of PQ and PR

5. If 7 sin2Ѳ + 3 cos2Ѳ = 4, show that tanѲ= 1/√3

6. If secɵ - tanɵ = 4, then prove that cosɵ = 8/17

7. If cosɵ - sinɵ = √2 sinɵ, prove that cosɵ + sinɵ = √2 cosɵ

8. If √3 tanѲ = 3 sinѲ, find the value of sin2Ѳ - cos2Ѳ

9. Evaluate: √2 tan245˚ + cos230˚ - sin260˚ (√2) 10. Evaluate: tan2 60˚ - 2 cos260˚ - ¾ sin2 45˚ - 4 sin2 30˚ (9/8)

11. Evaluate: (sin90˚ + cos45˚ + cos60˚)(cos0˚ - sin45˚ + sin30˚) (7/4)

12. If sin 2x = sin60˚cos30˚ - cos60˚ sin30˚, find x (15),

13. If A = B= 30•, verify that :

Sin(A + B ) = sin A cos B + cosA sinB

14. If sec2Ѳ (1+sinѲ) (1-sinѲ) = k, find the value of k (k = 1)

15. Evaluate: sec2 54˚ - cot236˚ + 2 sin238˚ sec2 52˚ - sin245˚

Cosec2 57˚ - tan233˚ (5/2)

16. Evaluate: sec (90 – Ѳ)cosecѲ – tan (90 – Ѳ)cotѲ + cos235 + cos255 (2)

Tan5˚ tan15˚ tan45˚ tan75˚ tan85˚

17. Find the value of:

2 sin 68˚ 2 cot 15˚ 3 tan45˚ tan20˚ tan40˚ tan50˚ tan70˚ (1)

Cos 22˚ 5 tan75˚ 5

18. If cos (40˚ + x) = sin 30˚, find the value of x (20˚)

19. Sin 4A = cos (A - 20˚), where 4A is an acute angle, find the value of A (22˚)

20. Find the value of Ѳ in 2 cos 3Ѳ = 1 ( 20˚)

21. Solve for Ѳ: 2 sin2Ѳ = ½ (30˚)

22. If sinѲ + cosѲ = √2cos (90˚ - Ѳ), determine cotѲ (√2 – 1)

23. Find the acute angles A and B, A>B, if sin (A + 2B) = √3/2 and cos (A + 4B) = 0 (30˚, 15˚)

24. If tan (A + B) = √3, tan (A – B) = 1, 0˚<A +B ≤ 90˚, a>b, then find A and B (52.5, 7.5)

25. If sin (A + B) = 1, cos (A – B) = 1, find A and B (45˚, 45˚)

26. If sinA – cosB = 0, prove that A + B = 90˚

27. What is the maximum value of 1/secѲ

28. Express cos56˚ + cot56˚ in terms of 0˚ and 45˚

29. Express cosA in terms of tanA

30. Find the value of tan 60˚ geometrically

31. If A, B and C are interior angles of triangle ABC, show that cos B+C = sin A

32. If x = a sinѲ, y = b tanѲ. Prove that a2 - b2 = 1 2 2

X2 y2

33. Prove that: 1 + 1 = 2 sec2 Ѳ

1 + sinѲ 1 – sinѲ

34. Prove that: sinѲ + 1 + cosѲ = 2cosecѲ

1 + cosѲ sinѲ

35. Prove: 1 + sin A = cosA

1 + sin A 1 – sinA

Question. A 1.6 m tall girl stands at a distance of 3.2m from a lamp post and casts a shadow of 4.8 m on the ground. Find the height of the lamp post
Answer : (2.6m)

Question. A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill is 60˚ and the angle of depression of the base of the hill is 30.˚ Calculate the distance of the hill from the ship and the height of the hill
Answer : (10√3m, 40m)

Question. The angle of elevation of a cloud from a point 60m above a lake is 30˚ and angle of depression of the reflection of cloud in the Lake is 60˚. Find the height of the cloud.
Answer : (120 m)

Question. The angle of elevation of a jet plane from a point A on the ground is 60˚. After a flight of 15 sec the angle of elevation changes to 30˚. If the jet plane is flying at a constant height of 1500√3m, then find the speed of jet plane.
Answer : (720 km /hr)

Question. A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane, the angles of elevation at the bottom and the top of the flagstaff are α and β respectively.
Answer : Prove that the height of the tower is h tan α / tanβ – tanα

Question. The angle of elevation of the top of a tower from two points at distances a and b metres from the base and in the same straight line with it are complementary.
Answer : Prove that height of the tower is √ab metres.

Question. The angles of elevation of the top of a rock from the top and foot of a 100 m high tower are 30˚ and 45˚respectively. Find the height of the rock.
Answer : (236.5 m)

Question. A boy is standing on the ground and is flying a kite with 100m of string at an elevation of 30˚ Another boy is standing on the roof of a 10m high building and is flying his kite at an elevation of 45˚. Both the boys are on opposite sides of the kite’s .Find the length of the string that the Second boy must have so that two kites meet.
Answer : (40√2 m)

Question. the shadow of a tower standing on a level ground is found to be 40 m longer when the sun, s altitude is 30˚ than when it is 60˚. Find the height of the tower.
Answer : (20√3m)

Question. The angle of elevation ø of a vertical tower from a point on ground is such that its tangent is 5/12. On walking 192m towards the tower in the same straight line, the tangent of the angle of elevation Is found to be ¾. Find the height of the tower
Answer : (180 m)

Question. A bird is sitting on the top of a tree, which is 80m high. The angle of elevation of the bird, from a point on the ground is 45˚. The bird flies away from the point of observation horizontally and remains at a Constant height. After 2 sec, the angle of Elevation of the bird from the point of observation becomes 30˚. Find the speed of flying of the bird
Answer : (29.28m/sec)

Question. An aero plane at an altitude of 200m observes the angles of depression of opposite points on the two banks of a river to be 45˚ and 60˚. Find the width of the river
Answer : (315.4m)

Question. Two men on either side of a cliff, 60m high, observe the angles of elevation of the top of the cliff to be 45˚ and 60˚ respectively Find the distance between two men
Answer : (94.6m)

Question. From the top of a tower the angle of depression of an object on the horizontal ground is found to be 60˚. On descending 20m Vertically downwards from the top of the tower, the angle of depression of the object is found to be 30˚. Find the height of the Tower.
Answer : (30 m)

Question. A pole 6 m high casts a shadow 2√3 m long on the ground, then the sun ,s elevation is
a) 60⁰
b) 45⁰
c) 30⁰
d) 90⁰

Question. If AB = 4 m and AC= 8 m , then angle of observation of A as observed from C is
a) 60⁰
b) 30⁰
c) 45⁰
d) cannot be determined

Question. When the sun is 30⁰ above the horizontal, the length of shadow cast by 50 m building is
a) 50/√3 m
b) 50 √3 m
c) 25 √3 m
d) none of these

Question. When the height of the shadow of a pole is equal to the height of the pole then the elevation of source of light is
a) 30⁰
b) 20 √3
c) 60⁰
d) 45⁰

Question. The angle formed by the line of sight with the horizontal, when the point being viewed is above the horizontal level is called
a) Vertical angle
b) angle of depression
c) angle of elevation
d) obtuse angle

Trigonometry Practice Sheet

Question 1. If \(\cot\Theta = 15/8\), evaluate \(\frac{(2 + 2\sin\Theta)(1 - \sin\Theta)}{(1 + \cos\Theta)(2 - 2\cos\Theta)}\)
Answer:
We are given \(\cot\Theta = \frac{15}{8}\).
The given expression is:
\[ E = \frac{(2 + 2\sin\Theta)(1 - \sin\Theta)}{(1 + \cos\Theta)(2 - 2\cos\Theta)} \]
Taking 2 common from the numerator and denominator:
\[ E = \frac{2(1 + \sin\Theta)(1 - \sin\Theta)}{2(1 + \cos\Theta)(1 - \cos\Theta)} \]
\[ E = \frac{1 - \sin^2\Theta}{1 - \cos^2\Theta} \]
Using the identity \(\sin^2\Theta + \cos^2\Theta = 1\), we have \(1 - \sin^2\Theta = \cos^2\Theta\) and \(1 - \cos^2\Theta = \sin^2\Theta\):
\[ E = \frac{\cos^2\Theta}{\sin^2\Theta} = \cot^2\Theta \]
Substituting the value of \(\cot\Theta\):
\[ E = \left(\frac{15}{8}\right)^2 = \frac{225}{64} \]
In simple words: Simplify the algebraic terms in the fraction first. After applying standard trigonometric identities, the expression reduces directly to the square of cotangent, which can be evaluated directly.

Exam Tip: Factoring out common numbers first simplifies the expression and avoids tedious calculations involving trigonometric ratios early in the problem.

Question 2. If \(\tan A = 2\) . Evaluate \(\sec A \sin A + \tan^2 A - \csc A\)
Answer:
Given \(\tan A = 2 = \frac{2}{1}\).
We know that \(\tan A = \frac{\text{Perpendicular (P)}}{\text{Base (B)}}\). Let \(P = 2k\) and \(B = 1k\).
Using Pythagoras theorem, \(\text{Hypotenuse (H)} = \sqrt{P^2 + B^2} = \sqrt{(2k)^2 + (1k)^2} = k\sqrt{5}\).
Now we can find the required trigonometric ratios:
\[ \sin A = \frac{P}{H} = \frac{2}{\sqrt{5}} \]
\[ \sec A = \frac{H}{B} = \sqrt{5} \]
\[ \csc A = \frac{H}{P} = \frac{\sqrt{5}}{2} \]
Substituting these values into the expression \(\sec A \sin A + \tan^2 A - \csc A\):
\[ \left(\sqrt{5}\right) \left(\frac{2}{\sqrt{5}}\right) + (2)^2 - \frac{\sqrt{5}}{2} \]
\[ = 2 + 4 - \frac{\sqrt{5}}{2} = 6 - \frac{\sqrt{5}}{2} = \frac{12 - \sqrt{5}}{2} \]
In simple words: Use the given tangent value to construct a right-angled triangle. Find the hypotenuse using the Pythagorean theorem, and then substitute the values of the other required trigonometric ratios into the expression.

Exam Tip: Notice that \(\sec A \sin A = \frac{1}{\cos A} \sin A = \tan A\). Simplifying parts of the expression algebraically beforehand can save computation time.

Question 3. In a \(\Delta ABC\), right angled at \(A\), if \(\tan C = \sqrt{3}\), find the value of \(\sin B \cos C + \cos B \sin C\)
Answer:
In a right-angled triangle \(ABC\) where \(\angle A = 90^\circ\):
The sum of angles in a triangle is \(180^\circ\), so \(A + B + C = 180^\circ \implies B + C = 90^\circ\).
Using the trigonometric addition formula:
\[ \sin B \cos C + \cos B \sin C = \sin(B + C) \]
Substituting \(B + C = 90^\circ\):
\[ \sin(90^\circ) = 1 \]
In simple words: Since the triangle is right-angled at A, the sum of the other two angles B and C must be 90 degrees. The given expression represents the expansion of sine of the sum of B and C, which equals sine of 90 degrees.

Exam Tip: Recognizing standard expansions like \(\sin(x + y)\) immediately bypasses the need to find individual side lengths or angles, making the solution much faster.

Question 4. In \(\Delta PQR\), right angled at \(Q\), \(QR = 6\text{ cm}\), \(\angle QPR = 60^\circ\). Find the length of \(PQ\) and \(PR\)
Answer:
In right-angled triangle \(PQR\) with \(\angle Q = 90^\circ\):
Using the tangent ratio for \(\angle QPR = 60^\circ\):
\[ \tan(60^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{QR}{PQ} \]
\[ \sqrt{3} = \frac{6}{PQ} \implies PQ = \frac{6}{\sqrt{3}} = 2\sqrt{3}\text{ cm} \]
Now, using the sine ratio to find \(PR\):
\[ \sin(60^\circ) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{QR}{PR} \]
\[ \frac{\sqrt{3}}{2} = \frac{6}{PR} \implies PR = \frac{12}{\sqrt{3}} = 4\sqrt{3}\text{ cm} \]
Thus, the lengths are \(PQ = 2\sqrt{3}\text{ cm}\) and \(PR = 4\sqrt{3}\text{ cm}\).
In simple words: Apply basic trigonometric ratios like tangent and sine to the given angle of 60 degrees. Use the known opposite side of 6 cm to compute the adjacent side and hypotenuse respectively.

Exam Tip: Rationalize the denominator when dealing with surds in the denominator (like multiplying top and bottom by \(\sqrt{3}\)) to write the final answers in standard simplified form.

Question 5. If \(7\sin^2\Theta + 3\cos^2\Theta = 4\), show that \(\tan\Theta = 1/\sqrt{3}\)
Answer:
We are given the equation:
\[ 7\sin^2\Theta + 3\cos^2\Theta = 4 \]
We can split \(7\sin^2\Theta\) as \(4\sin^2\Theta + 3\sin^2\Theta\):
\[ 4\sin^2\Theta + 3\sin^2\Theta + 3\cos^2\Theta = 4 \]
\[ 4\sin^2\Theta + 3(\sin^2\Theta + \cos^2\Theta) = 4 \]
Since \(\sin^2\Theta + \cos^2\Theta = 1\):
\[ 4\sin^2\Theta + 3(1) = 4 \]
\[ 4\sin^2\Theta = 1 \implies \sin^2\Theta = \frac{1}{4} \]
Taking the positive square root for an acute angle:
\[ \sin\Theta = \frac{1}{2} \]
For \(\sin\Theta = \frac{1}{2}\), the acute angle is \(\Theta = 30^\circ\).
Thus, \(\tan\Theta = \tan(30^\circ) = \frac{1}{\sqrt{3}}\).
In simple words: Break down the terms to utilize the fundamental identity \(\sin^2\Theta + \cos^2\Theta = 1\). This reduces the equation to a single trigonometric term from which the angle and tangent value can be computed.

Exam Tip: Splitting terms to isolate standard identities is a highly effective algebraic tool in trigonometry that simplifies equations quickly.

Question 6. If \(\sec\Theta - \tan\Theta = 4\), then prove that \(\cos\Theta = 8/17\)
Answer:
We know the identity:
\[ \sec^2\Theta - \tan^2\Theta = 1 \]
We can write this as:
\[ (\sec\Theta - \tan\Theta)(\sec\Theta + \tan\Theta) = 1 \]
Given that \(\sec\Theta - \tan\Theta = 4\), we substitute this in:
\[ 4(\sec\Theta + \tan\Theta) = 1 \implies \sec\Theta + \tan\Theta = \frac{1}{4} \]
Let \(\sec\Theta - \tan\Theta = 4\) be equation (1) and \(\sec\Theta + \tan\Theta = 1/4\) be equation (2). Adding both equations:
\[ 2\sec\Theta = 4 + \frac{1}{4} = \frac{17}{4} \]
\[ \sec\Theta = \frac{17}{8} \]
Since \(\cos\Theta = \frac{1}{\sec\Theta}\):
\[ \cos\Theta = \frac{8}{17} \]
Hence proved.
In simple words: Use the standard identity connecting secant and tangent to establish a system of linear equations. Solving these equations yields the value of secant, whose reciprocal is the required cosine value.

Exam Tip: Remember that if \(\sec\Theta - \tan\Theta = x\), then \(\sec\Theta + \tan\Theta = 1/x\). This relationship is frequently used in board exam questions.

Question 7. If \(\cos\Theta - \sin\Theta = \sqrt{2}\sin\Theta\), prove that \(\cos\Theta + \sin\Theta = \sqrt{2}\cos\Theta\)
Answer:
Given:
\[ \cos\Theta - \sin\Theta = \sqrt{2}\sin\Theta \]
Rearranging the terms:
\[ \cos\Theta = \sqrt{2}\sin\Theta + \sin\Theta = (\sqrt{2} + 1)\sin\Theta \]
Multiplying both sides by \((\sqrt{2} - 1)\):
\[ (\sqrt{2} - 1)\cos\Theta = (\sqrt{2} - 1)(\sqrt{2} + 1)\sin\Theta \]
\[ (\sqrt{2} - 1)\cos\Theta = (2 - 1)\sin\Theta = \sin\Theta \]
Expanding the left side:
\[ \sqrt{2}\cos\Theta - \cos\Theta = \sin\Theta \]
Rearranging the terms to get the desired form:
\[ \cos\Theta + \sin\Theta = \sqrt{2}\cos\Theta \]
Hence proved.
In simple words: Group the sine terms together and isolate cosine. Multiplying by the conjugate of the coefficient rationalizes the expression and directly leads to the required identity.

Exam Tip: Alternatively, squaring both sides of the given equation and substituting \(\cos^2\Theta = 1 - \sin^2\Theta\) is another mathematically sound approach to arrive at the proof.

Question 8. If \(\sqrt{3}\tan\Theta = 3\sin\Theta\), find the value of \(\sin^2\Theta - \cos^2\Theta\)
Answer:
Given:
\[ \sqrt{3}\tan\Theta = 3\sin\Theta \]
Since \(\tan\Theta = \frac{\sin\Theta}{\cos\Theta}\):
\[ \sqrt{3}\frac{\sin\Theta}{\cos\Theta} = 3\sin\Theta \]
Assuming \(\sin\Theta \neq 0\), we can divide both sides by \(\sin\Theta\):
\[ \frac{\sqrt{3}}{\cos\Theta} = 3 \implies \cos\Theta = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \]
Now, we need to find \(\sin^2\Theta - \cos^2\Theta\):
Using the identity \(\sin^2\Theta = 1 - \cos^2\Theta\):
\[ \sin^2\Theta - \cos^2\Theta = (1 - \cos^2\Theta) - \cos^2\Theta = 1 - 2\cos^2\Theta \]
Substituting the value of \(\cos\Theta\):
\[ 1 - 2\left(\frac{1}{\sqrt{3}}\right)^2 = 1 - 2\left(\frac{1}{3}\right) = 1 - \frac{2}{3} = \frac{1}{3} \]
In simple words: Express tangent in terms of sine and cosine to simplify the equation and find the value of cosine. Use this cosine value to evaluate the requested difference of squares.

Exam Tip: Substituting the identity to convert the target expression entirely into a single trigonometric ratio (like cosine) saves you from calculating the sine value separately.

Question 9. Evaluate: \(\sqrt{2}\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ\)
Answer:
Using the standard values of trigonometric ratios:
- \(\tan 45^\circ = 1\)
- \(\cos 30^\circ = \frac{\sqrt{3}}{2}\)
- \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)
Substituting these values into the given expression:
\[ E = \sqrt{2}(\tan 45^\circ)^2 + (\cos 30^\circ)^2 - (\sin 60^\circ)^2 \]
\[ E = \sqrt{2}(1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2 \]
The last two terms cancel out, leaving:
\[ E = \sqrt{2} \]
In simple words: Plug in the standard exact numerical values for the trigonometric ratios of 45, 30, and 60 degrees to find the simplified final numerical value.

Exam Tip: Recognizing that \(\cos 30^\circ = \sin 60^\circ\) allows you to cancel their squared terms instantly without squaring the fractions.

Question 10. Evaluate: \(\tan^2 60^\circ - 2\cos^2 60^\circ - \frac{3}{4}\sin^2 45^\circ - 4\sin^2 30^\circ\)
Answer:
Substitute the standard trigonometric values:
- \(\tan 60^\circ = \sqrt{3}\)
- \(\cos 60^\circ = \frac{1}{2}\)
- \(\sin 45^\circ = \frac{1}{\sqrt{2}}\)
- \(\sin 30^\circ = \frac{1}{2}\)
Substitute these values into the expression:
\[ E = (\sqrt{3})^2 - 2\left(\frac{1}{2}\right)^2 - \frac{3}{4}\left(\frac{1}{\sqrt{2}}\right)^2 - 4\left(\frac{1}{2}\right)^2 \]
\[ E = 3 - 2\left(\frac{1}{4}\right) - \frac{3}{4}\left(\frac{1}{2}\right) - 4\left(\frac{1}{4}\right) \]
\[ E = 3 - \frac{1}{2} - \frac{3}{8} - 1 \]
\[ E = 2 - \frac{4}{8} - \frac{3}{8} = 2 - \frac{7}{8} = \frac{9}{8} \]
In simple words: Substitute the exact trigonometric ratios for each term, square them carefully, and combine the resulting fractions to find the answer.

Exam Tip: Work with a common denominator (in this case, 8) when performing the final fractional subtraction to avoid simple arithmetic mistakes.

Question 11. Evaluate: \((\sin 90^\circ + \cos 45^\circ + \cos 60^\circ)(\cos 0^\circ - \sin 45^\circ + \sin 30^\circ)\)
Answer:
Substitute the known trigonometric values:
- \(\sin 90^\circ = 1\), \(\cos 45^\circ = \frac{1}{\sqrt{2}}\), \(\cos 60^\circ = \frac{1}{2}\)
- \(\cos 0^\circ = 1\), \(\sin 45^\circ = \frac{1}{\sqrt{2}}\), \(\sin 30^\circ = \frac{1}{2}\)
Substituting these values into the given expression:
\[ E = \left(1 + \frac{1}{\sqrt{2}} + \frac{1}{2}\right)\left(1 - \frac{1}{\sqrt{2}} + \frac{1}{2}\right) \]
Combine the rational numbers inside each bracket:
\[ E = \left(\frac{3}{2} + \frac{1}{\sqrt{2}}\right)\left(\frac{3}{2} - \frac{1}{\sqrt{2}}\right) \]
Using the algebraic identity \((a + b)(a - b) = a^2 - b^2\):
\[ E = \left(\frac{3}{2}\right)^2 - \left(\frac{1}{\sqrt{2}}\right)^2 \]
\[ E = \frac{9}{4} - \frac{1}{2} = \frac{9}{4} - \frac{2}{4} = \frac{7}{4} \]
In simple words: Group the numerical values to form a product of conjugate pairs. Applying the difference of squares identity simplifies the calculation significantly.

Exam Tip: Grouping terms (rational together and irrational together) before expanding prevents long, complicated multiplications of surds.

Question 12. If \(\sin 2x = \sin 60^\circ\cos 30^\circ - \cos 60^\circ\sin 30^\circ\), find \(x\)
Answer:
The right-hand side is of the form \(\sin A \cos B - \cos A \sin B\), which is equal to \(\sin(A - B)\).
Thus, the given equation becomes:
\[ \sin 2x = \sin(60^\circ - 30^\circ) \]
\[ \sin 2x = \sin 30^\circ \]
Comparing the angles on both sides:
\[ 2x = 30^\circ \implies x = 15^\circ \]
In simple words: Recognize that the right side of the equation corresponds to the sine subtraction formula. This simplifies the equation to a direct trigonometric comparison.

Exam Tip: If you do not remember the identity, you can also substitute the standard numerical values directly on the right side to get \(\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{3}{4} - \frac{1}{4} = \frac{1}{2} = \sin 30^\circ\).

Question 13. If \(A = B = 30^\circ\), verify that : \(\sin(A + B) = \sin A\cos B + \cos A\sin B\)
Answer:
Given \(A = B = 30^\circ\).
Evaluating the Left-Hand Side (LHS):
\[ \text{LHS} = \sin(A + B) = \sin(30^\circ + 30^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
Evaluating the Right-Hand Side (RHS):
\[ \text{RHS} = \sin A \cos B + \cos A \sin B = \sin 30^\circ \cos 30^\circ + \cos 30^\circ \sin 30^\circ \]
Substitute standard values \(\sin 30^\circ = \frac{1}{2}\) and \(\cos 30^\circ = \frac{\sqrt{3}}{2}\):
\[ \text{RHS} = \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) \]
\[ \text{RHS} = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \]
Since \(\text{LHS} = \text{RHS}\), the relation is verified.
In simple words: Compute both sides of the equation independently by plugging in 30 degrees for both angles. Show that both calculations lead to the exact same final result.

Exam Tip: Always clearly structure verification problems by showing LHS and RHS computations separately before concluding they are equal.

Question 14. If \(\sec^2\Theta (1+\sin\Theta) (1-\sin\Theta) = k\), find the value of \(k\)
Answer:
Given expression:
\[ \sec^2\Theta (1+\sin\Theta) (1-\sin\Theta) = k \]
Using the algebraic identity \((1 + a)(1 - a) = 1 - a^2\):
\[ \sec^2\Theta (1 - \sin^2\Theta) = k \]
Using the identity \(1 - \sin^2\Theta = \cos^2\Theta\):
\[ \sec^2\Theta \cdot \cos^2\Theta = k \]
Since \(\sec\Theta \cdot \cos\Theta = 1\):
\[ (\sec\Theta \cdot \cos\Theta)^2 = k \implies (1)^2 = k \implies k = 1 \]
Thus, the value of \(k\) is 1.
In simple words: Combine the binomial factors into a single squared term using basic algebra, then use fundamental trigonometric identities to cancel out the terms.

Exam Tip: Keeping identities like \(\sec^2\Theta - \tan^2\Theta = 1\) and \(\sin^2\Theta + \cos^2\Theta = 1\) in mind is essential for simplifying product-based questions.

Question 15. Evaluate: \(\frac{\sec^2 54^\circ - \cot^2 36^\circ}{\csc^2 57^\circ - \tan^2 33^\circ} + 2\sin^2 38^\circ \sec^2 52^\circ - \sin^2 45^\circ\)
Answer:
We use complementary angle relations:
- \(\cot 36^\circ = \tan(90^\circ - 36^\circ) = \tan 54^\circ\)
- \(\tan 33^\circ = \cot(90^\circ - 33^\circ) = \cot 57^\circ\)
- \(\sec 52^\circ = \csc(90^\circ - 52^\circ) = \csc 38^\circ\)
Now, rewrite the terms in the expression:
Numerator of the first term: \(\sec^2 54^\circ - \tan^2 54^\circ = 1\) (Using \(\sec^2\Theta - \tan^2\Theta = 1\))
Denominator of the first term: \(\csc^2 57^\circ - \cot^2 57^\circ = 1\) (Using \(\csc^2\Theta - \cot^2\Theta = 1\))
So, the first term becomes \(\frac{1}{1} = 1\).
The second term is:
\[ 2\sin^2 38^\circ \sec^2 52^\circ = 2\sin^2 38^\circ \csc^2 38^\circ \]
Since \(\sin\Theta \cdot \csc\Theta = 1\):
\[ 2(\sin 38^\circ \csc 38^\circ)^2 = 2(1)^2 = 2 \]
The third term is \(\sin^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}\).
Combine all terms:
\[ E = 1 + 2 - \frac{1}{2} = 3 - \frac{1}{2} = \frac{5}{2} \]
In simple words: Convert angles using complementary identities to obtain common angles. This allows you to apply standard pythagorean and reciprocal identities.

Exam Tip: Look for angles that sum to 90 degrees (such as 54 and 36, or 57 and 33) to identify which complementary conversions to make.

Question 16. Evaluate: \(\frac{\sec (90 - \Theta)\csc\Theta - \tan (90 - \Theta)\cot\Theta + \cos^2 35^\circ + \cos^2 55^\circ}{\tan 5^\circ \tan 15^\circ \tan 45^\circ \tan 75^\circ \tan 85^\circ}\)
Answer:
Let's simplify the numerator first:
Using complementary angle formulas:
- \(\sec(90^\circ - \Theta) = \csc\Theta\)
- \(\tan(90^\circ - \Theta) = \cot\Theta\)
- \(\cos 55^\circ = \sin(90^\circ - 55^\circ) = \sin 35^\circ\)
So, the numerator becomes:
\[ \csc\Theta \cdot \csc\Theta - \cot\Theta \cdot \cot\Theta + \cos^2 35^\circ + \sin^2 35^\circ \]
\[ = (\csc^2\Theta - \cot^2\Theta) + (\cos^2 35^\circ + \sin^2 35^\circ) \]
Using the identities \(\csc^2\Theta - \cot^2\Theta = 1\) and \(\sin^2 A + \cos^2 A = 1\):
\[ = 1 + 1 = 2 \]
Now simplify the denominator:
Using \(\tan(90^\circ - x) = \cot x\):
- \(\tan 85^\circ = \cot 5^\circ\)
- \(\tan 75^\circ = \cot 15^\circ\)
- \(\tan 45^\circ = 1\)
So, the denominator is:
\[ \tan 5^\circ \cdot \tan 15^\circ \cdot (1) \cdot \cot 15^\circ \cdot \cot 5^\circ \]
Rearranging terms:
\[ = (\tan 5^\circ \cdot \cot 5^\circ)(\tan 15^\circ \cdot \cot 15^\circ) = 1 \cdot 1 = 1 \]
Evaluating the entire fraction:
\[ E = \frac{\text{Numerator}}{\text{Denominator}} = \frac{2}{1} = 2 \]
In simple words: Convert the complementary angles in both numerator and denominator. This simplifies the top to the sum of basic identities, and the bottom to a product of reciprocals that cancel to 1.

Exam Tip: Be methodical and handle the numerator and denominator as separate sub-problems to keep your workspace clear and prevent errors.

Question 17. Find the value of: \(\frac{2\sin 68^\circ}{\cos 22^\circ} - \frac{2\cot 15^\circ}{5\tan 75^\circ} - \frac{3\tan 45^\circ \tan 20^\circ \tan 40^\circ \tan 50^\circ \tan 70^\circ}{5}\)
Answer:
We simplify each of the three terms separately using complementary angles:
1. First term: \(\cos 22^\circ = \sin(90^\circ - 22^\circ) = \sin 68^\circ\)
\[ \frac{2\sin 68^\circ}{\sin 68^\circ} = 2 \]
2. Second term: \(\tan 75^\circ = \cot(90^\circ - 75^\circ) = \cot 15^\circ\)
\[ \frac{2\cot 15^\circ}{5\cot 15^\circ} = \frac{2}{5} \]
3. Third term: \(\tan 70^\circ = \cot 20^\circ\), \(\tan 50^\circ = \cot 40^\circ\), and \(\tan 45^\circ = 1\)
\[ \frac{3(1) \tan 20^\circ \tan 40^\circ \cot 40^\circ \cot 20^\circ}{5} \]
Since \(\tan \theta \cdot \cot \theta = 1\):
\[ = \frac{3 \cdot 1 \cdot 1}{5} = \frac{3}{5} \]
Now, combining all three simplified terms:
\[ E = 2 - \frac{2}{5} - \frac{3}{5} = 2 - \left(\frac{2+3}{5}\right) = 2 - 1 = 1 \]
In simple words: Convert the ratios with complementary angles to find common terms that cancel out. Combining the simplified values yields the final result of 1.

Exam Tip: Watch the signs (plus or minus) between terms during fractional calculations to make sure you do not make simple arithmetic slips.

Question 18. If \(\cos (40^\circ + x) = \sin 30^\circ\), find the value of x
Answer:
Given:
\[ \cos(40^\circ + x) = \sin 30^\circ \]
We know that \(\sin 30^\circ = \frac{1}{2}\). Also, \(\cos 60^\circ = \frac{1}{2}\).
So, we can write:
\[ \cos(40^\circ + x) = \cos 60^\circ \]
Equating the angles:
\[ 40^\circ + x = 60^\circ \]
\[ x = 60^\circ - 40^\circ = 20^\circ \]
In simple words: Find the value of sine of 30 degrees and rewrite it as cosine of 60 degrees. This lets you compare the angles directly to solve for x.

Exam Tip: Alternatively, use the complementary angle formula directly: \(\sin 30^\circ = \cos(90^\circ - 30^\circ) = \cos 60^\circ\).

Question 19. Sin 4A = cos (A - 20^\circ), where 4A is an acute angle, find the value of A
Answer:
Given:
\[ \sin 4A = \cos(A - 20^\circ) \]
Using the complementary relation \(\sin\theta = \cos(90^\circ - \theta)\):
\[ \cos(90^\circ - 4A) = \cos(A - 20^\circ) \]
Equating the angles:
\[ 90^\circ - 4A = A - 20^\circ \]
\[ 90^\circ + 20^\circ = A + 4A \]
\[ 110^\circ = 5A \]
\[ A = \frac{110^\circ}{5} = 22^\circ \]
In simple words: Rewrite sine on the left side of the equation as cosine of its complement. Equating both sides allows you to solve a linear equation for A.

Exam Tip: Always verify that your final answer satisfies any given condition, such as \(4A\) being an acute angle (\(4 \times 22^\circ = 88^\circ < 90^\circ\)).

Question 20. Find the value of \(\Theta\) in \(2 \cos 3\Theta = 1\)
Answer:
Given equation:
\[ 2\cos 3\Theta = 1 \]
\[ \cos 3\Theta = \frac{1}{2} \]
We know that \(\cos 60^\circ = \frac{1}{2}\). Therefore:
\[ \cos 3\Theta = \cos 60^\circ \]
Equating the angles:
\[ 3\Theta = 60^\circ \]
\[ \Theta = 20^\circ \]
In simple words: Isolate the cosine term on one side of the equation. Match it with a standard angle value to find the unknown angle.

Exam Tip: Be careful not to divide the angle inside the cosine function before evaluating the value of the function.

Question 21. Solve for \(\Theta\): \(2 \sin^2\Theta = 1/2\)
Answer:
Given:
\[ 2\sin^2\Theta = \frac{1}{2} \]
Divide both sides by 2:
\[ \sin^2\Theta = \frac{1}{4} \]
Taking square root on both sides (considering positive value for acute angle):
\[ \sin\Theta = \frac{1}{2} \]
We know that \(\sin 30^\circ = \frac{1}{2}\). Therefore:
\[ \Theta = 30^\circ \]
In simple words: Isolate the squared sine term, take the square root, and identify which standard angle corresponds to that value.

Exam Tip: Always assume the angle is acute unless specified otherwise, which means we focus on positive values of trigonometric ratios.

Question 22. If \(\sin\Theta + \cos\Theta = \sqrt{2}\cos(90^\circ - \Theta)\), determine \(\cot\Theta\)
Answer:
Given equation:
\[ \sin\Theta + \cos\Theta = \sqrt{2}\cos(90^\circ - \Theta) \]
Using the identity \(\cos(90^\circ - \Theta) = \sin\Theta\):
\[ \sin\Theta + \cos\Theta = \sqrt{2}\sin\Theta \]
Divide the entire equation by \(\sin\Theta\):
\[ \frac{\sin\Theta}{\sin\Theta} + \frac{\cos\Theta}{\sin\Theta} = \frac{\sqrt{2}\sin\Theta}{\sin\Theta} \]
\[ 1 + \cot\Theta = \sqrt{2} \]
\[ \cot\Theta = \sqrt{2} - 1 \]
In simple words: First simplify the right-hand side using complementary relations. Dividing the entire equation by sine converts the expression directly to cotangent, allowing you to solve for it.

Exam Tip: Dividing an equation containing sine and cosine by sine or cosine is an excellent shortcut to introduce tangent or cotangent directly.

Question 23. Find the acute angles A and B, A>B, if \(\sin(A + 2B) = \sqrt{3}/2\) and \(\cos(A + 4B) = 0\)
Answer:
We are given:
\[ \sin(A + 2B) = \frac{\sqrt{3}}{2} \]
Since \(\sin 60^\circ = \frac{\sqrt{3}}{2}\):
\[ A + 2B = 60^\circ \quad \text{--- (Equation 1)} \]
We are also given:
\[ \cos(A + 4B) = 0 \]
Since \(\cos 90^\circ = 0\):
\[ A + 4B = 90^\circ \quad \text{--- (Equation 2)} \]
Subtract Equation 1 from Equation 2:
\[ (A + 4B) - (A + 2B) = 90^\circ - 60^\circ \]
\[ 2B = 30^\circ \implies B = 15^\circ \]
Substitute \(B = 15^\circ\) back into Equation 1:
\[ A + 2(15^\circ) = 60^\circ \]
\[ A + 30^\circ = 60^\circ \implies A = 30^\circ \]
So, \(A = 30^\circ\) and \(B = 15^\circ\). Since \(30^\circ > 15^\circ\), the condition \(A > B\) is satisfied.
In simple words: Convert the given trigonometric equations into two linear equations by matching them with standard values. Solve these simultaneous equations to find the angles.

Exam Tip: Elimination is the simplest way to solve linear systems when the coefficients of one variable (like \(A\)) match perfectly.

Question 24. If \(\tan(A + B) = \sqrt{3}\), \(\tan(A - B) = 1\), \(0^\circ<A +B \le 90^\circ\), \(a>b\), then find A and B
Answer:
Given equations:
1. \(\tan(A + B) = \sqrt{3}\)
Since \(\tan 60^\circ = \sqrt{3}\):
\[ A + B = 60^\circ \quad \text{--- (Equation 1)} \]
2. \(\tan(A - B) = 1\)
Since \(\tan 45^\circ = 1\):
\[ A - B = 45^\circ \quad \text{--- (Equation 2)} \]
Adding Equation 1 and Equation 2:
\[ (A + B) + (A - B) = 60^\circ + 45^\circ \]
\[ 2A = 105^\circ \implies A = 52.5^\circ \]
Subtracting Equation 2 from Equation 1:
\[ (A + B) - (A - B) = 60^\circ - 45^\circ \]
\[ 2B = 15^\circ \implies B = 7.5^\circ \]
Thus, \(A = 52.5^\circ\) and \(B = 7.5^\circ\).
In simple words: Convert the tangent equations into algebraic equations and solve them simultaneously to find the values of \(A\) and \(B\).

Exam Tip: Do not worry if you get fractional or decimal values for angles, as long as they satisfy the given inequality constraints.

Question 25. If \(\sin(A + B) = 1\), \(\cos(A - B) = 1\), find A and B
Answer:
We are given:
\[ \sin(A + B) = 1 \]
Since \(\sin 90^\circ = 1\):
\[ A + B = 90^\circ \quad \text{--- (Equation 1)} \]
We also have:
\[ \cos(A - B) = 1 \]
Since \(\cos 0^\circ = 1\):
\[ A - B = 0^\circ \quad \text{--- (Equation 2)} \]
From Equation 2, we have \(A = B\). Substituting this in Equation 1:
\[ 2A = 90^\circ \implies A = 45^\circ \]
Since \(A = B\), we have \(B = 45^\circ\).
Thus, \(A = 45^\circ\) and \(B = 45^\circ\).
In simple words: Set up algebraic equations from the trigonometric ratios. Since A minus B is 0, the two angles are equal, meaning each is half of 90 degrees.

Exam Tip: If the cosine of a difference is 1, it implies the two angles are identical for acute angles.

Question 26. If \(\sin A - \cos B = 0\), prove that \(A + B = 90^\circ\)
Answer:
Given:
\[ \sin A - \cos B = 0 \]
\[ \sin A = \cos B \]
We know the complementary identity \(\cos B = \sin(90^\circ - B)\). Substituting this in:
\[ \sin A = \sin(90^\circ - B) \]
Comparing the angles on both sides:
\[ A = 90^\circ - B \]
\[ A + B = 90^\circ \]
Hence proved.
In simple words: Equate sine and cosine, then express cosine of B in terms of its complementary sine angle. Equating the angles yields the proof.

Exam Tip: Complementary angle formulas are incredibly useful for transforming different trigonometric functions into a single type.

Question 27. What is the maximum value of \(1/\sec\Theta\)
Answer:
We know that:
\[ \frac{1}{\sec\Theta} = \cos\Theta \]
The value of \(\cos\Theta\) for any real angle \(\Theta\) lies in the range \([-1, 1]\).
Therefore, the maximum value of \(\cos\Theta\) is 1.
Thus, the maximum value of \(\frac{1}{\sec\Theta}\) is 1.
In simple words: The reciprocal of secant is cosine. Since the highest value cosine can ever reach is 1, that is also the maximum value for the given term.

Exam Tip: Always simplify reciprocal functions into primary ratios (sine, cosine, tangent) to determine range, boundaries, or limits easily.

Question 28. Express \(\cos 56^\circ + \cot 56^\circ\) in terms of \(0^\circ\) and \(45^\circ\)
Answer:
We use complementary angle identities:
- \(\cos\theta = \sin(90^\circ - \theta)\)
- \(\cot\theta = \tan(90^\circ - \theta)\)
Applying these to the given angles:
\[ \cos 56^\circ = \sin(90^\circ - 56^\circ) = \sin 34^\circ \]
\[ \cot 56^\circ = \tan(90^\circ - 56^\circ) = \tan 34^\circ \]
Therefore:
\[ \cos 56^\circ + \cot 56^\circ = \sin 34^\circ + \tan 34^\circ \]
Since \(34^\circ\) lies between \(0^\circ\) and \(45^\circ\), the expression is represented in the required range.
In simple words: Convert the given trigonometric terms to their complementary ratios to get angles that lie between 0 and 45 degrees.

Exam Tip: Be sure to compute the difference from 90 degrees correctly to prevent basic arithmetic subtraction errors.

Question 29. Express \(\cos A\) in terms of \(\tan A\)
Answer:
We know that:
\[ \cos A = \frac{1}{\sec A} \]
Using the identity \(\sec^2 A = 1 + \tan^2 A\), we have:
\[ \sec A = \sqrt{1 + \tan^2 A} \]
Substituting this back into the relation for \(\cos A\):
\[ \cos A = \frac{1}{\sqrt{1 + \tan^2 A}} \]
In simple words: Relate cosine to secant, then use the standard identity connecting secant to tangent to write the final formula.

Exam Tip: Using algebraic paths through the reciprocal identity is the easiest way to express any trigonometric function in terms of another.

Question 30. Find the value of \(\tan 60^\circ\) geometrically
Answer:
Consider an equilateral triangle \(ABC\) of side \(2a\). Since it is equilateral, each angle is \(60^\circ\).
Draw a perpendicular line \(AD\) from \(A\) to \(BC\). In an equilateral triangle, this altitude bisects the base and the vertical angle.
Therefore, \(BD = DC = a\), and \(\angle BAD = 30^\circ\), \(\angle ABD = 60^\circ\).
In right-angled triangle \(ABD\), using Pythagoras theorem:
\[ AD^2 + BD^2 = AB^2 \]
\[ AD^2 + a^2 = (2a)^2 \]
\[ AD^2 + a^2 = 4a^2 \implies AD^2 = 3a^2 \]
\[ AD = a\sqrt{3} \]
Now, in right triangle \(ABD\), for angle \(\angle ABD = 60^\circ\):
\[ \tan(60^\circ) = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AD}{BD} \]
\[ \tan(60^\circ) = \frac{a\sqrt{3}}{a} = \sqrt{3} \]
In simple words: Draw an equilateral triangle and bisect it with an altitude to get a 30-60-90 right triangle. Using Pythagoras theorem, calculate the height and find the ratio of height to base to get the value of tangent 60 degrees.

Exam Tip: Drawing a neat, labeled diagram of the bisected equilateral triangle is essential to secure full marks in geometric derivation questions.

Question 31. If A, B and C are interior angles of triangle ABC, show that \(\cos\left\{\frac{B+C}{2}\right\} = \sin\frac{A}{2}\)
Answer:
Since \(A, B,\) and \(C\) are interior angles of a triangle \(ABC\):
\[ A + B + C = 180^\circ \]
\[ B + C = 180^\circ - A \]
Dividing both sides by 2:
\[ \frac{B + C}{2} = 90^\circ - \frac{A}{2} \]
Taking cosine on both sides:
\[ \cos\left(\frac{B + C}{2}\right) = \cos\left(90^\circ - \frac{A}{2}\right) \]
Using the complementary identity \(\cos(90^\circ - \theta) = \sin\theta\):
\[ \cos\left(\frac{B + C}{2}\right) = \sin\left(\frac{A}{2}\right) \]
Hence proved.
In simple words: Start with the angle sum property of a triangle. Manipulate the equation by dividing by 2 and taking cosine on both sides to arrive at the desired result using complementary relations.

Exam Tip: This type of question is very common. Always start from \(A+B+C = 180^\circ\) and rearrange based on the groupings shown in the target expression.

Question 32. If \(x = a \sin\Theta\), \(y = b \tan\Theta\). Prove that \(\frac{a^2}{x^2} - \frac{b^2}{y^2} = 1\)
Answer:
We are given:
\[ x = a\sin\Theta \implies \frac{x}{a} = \sin\Theta \implies \frac{a}{x} = \frac{1}{\sin\Theta} = \csc\Theta \]
We also have:
\[ y = b\tan\Theta \implies \frac{y}{b} = \tan\Theta \implies \frac{b}{y} = \frac{1}{\tan\Theta} = \cot\Theta \]
Now, evaluate the left-hand side of the target equation:
\[ \text{LHS} = \frac{a^2}{x^2} - \frac{b^2}{y^2} = \left(\frac{a}{x}\right)^2 - \left(\frac{b}{y}\right)^2 \]
Substituting the values of \(\frac{a}{x}\) and \(\frac{b}{y}\):
\[ \text{LHS} = \csc^2\Theta - \cot^2\Theta \]
Using the fundamental identity \(\csc^2\Theta - \cot^2\Theta = 1\):
\[ \text{LHS} = 1 = \text{RHS} \]
Hence proved.
In simple words: Rearrange the equations to find the values of \(a/x\) and \(b/y\). Substituting these values into the expression yields a standard identity which equals 1.

Exam Tip: Isolating the trigonometric ratios before squaring makes substitution and applying trigonometric identities straightforward.

Question 33. Prove that: \(\frac{1}{1 + \sin\Theta} + \frac{1}{1 - \sin\Theta} = 2 \sec^2\Theta\)
Answer:
Let's evaluate the Left-Hand Side (LHS):
\[ \text{LHS} = \frac{1}{1 + \sin\Theta} + \frac{1}{1 - \sin\Theta} \]
Taking the common denominator:
\[ \text{LHS} = \frac{(1 - \sin\Theta) + (1 + \sin\Theta)}{(1 + \sin\Theta)(1 - \sin\Theta)} \]
In the numerator, \(\sin\Theta\) cancels out:
\[ \text{LHS} = \frac{2}{(1 - \sin^2\Theta)} \]
Using the identity \(1 - \sin^2\Theta = \cos^2\Theta\):
\[ \text{LHS} = \frac{2}{\cos^2\Theta} = 2\sec^2\Theta = \text{RHS} \]
Hence proved.
In simple words: Find a common denominator to add the fractions, simplify the numerator, and convert the resulting cosine term in the denominator to secant.

Exam Tip: Combine fractions in rational trigonometric proofs as your very first step; the simplified numerator often reveals a key identity.

Question 34. Prove that: \(\frac{\sin\Theta}{1 + \cos\Theta} + \frac{1 + \cos\Theta}{\sin\Theta} = 2\csc\Theta\)
Answer:
Let's simplify the Left-Hand Side (LHS):
\[ \text{LHS} = \frac{\sin\Theta}{1 + \cos\Theta} + \frac{1 + \cos\Theta}{\sin\Theta} \]
Find a common denominator:
\[ \text{LHS} = \frac{\sin^2\Theta + (1 + \cos\Theta)^2}{\sin\Theta(1 + \cos\Theta)} \]
Expand the term \((1 + \cos\Theta)^2\):
\[ \text{LHS} = \frac{\sin^2\Theta + 1 + \cos^2\Theta + 2\cos\Theta}{\sin\Theta(1 + \cos\Theta)} \]
Group \(\sin^2\Theta + \cos^2\Theta = 1\):
\[ \text{LHS} = \frac{1 + 1 + 2\cos\Theta}{\sin\Theta(1 + \cos\Theta)} \]
\[ \text{LHS} = \frac{2 + 2\cos\Theta}{\sin\Theta(1 + \cos\Theta)} \]
Factor out 2 in the numerator:
\[ \text{LHS} = \frac{2(1 + \cos\Theta)}{\sin\Theta(1 + \cos\Theta)} \]
Cancel the common term \((1 + \cos\Theta)\):
\[ \text{LHS} = \frac{2}{\sin\Theta} = 2\csc\Theta = \text{RHS} \]
Hence proved.
In simple words: Cross-multiply to add the fractions. Applying the identity \(\sin^2\Theta + \cos^2\Theta = 1\) simplifies the numerator, allowing the binomial term to cancel out, leaving cosecant.

Exam Tip: When expanding algebraic squares like \((1+\cos\theta)^2\), remember to include the middle term \(2\cos\theta\) to avoid incomplete expansions.

Question 35. Prove: \(\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \frac{\cos A}{1 - \sin A}\)
Answer:
Let's simplify the Left-Hand Side (LHS):
\[ \text{LHS} = \sqrt{\frac{1 + \sin A}{1 - \sin A}} \]
Multiply the numerator and denominator inside the square root by \((1 - \sin A)\):
\[ \text{LHS} = \sqrt{\frac{(1 + \sin A)(1 - \sin A)}{(1 - \sin A)(1 - \sin A)}} \]
\[ \text{LHS} = \sqrt{\frac{1 - \sin^2 A}{(1 - \sin A)^2}} \]
Using the identity \(1 - \sin^2 A = \cos^2 A\):
\[ \text{LHS} = \sqrt{\frac{\cos^2 A}{(1 - \sin A)^2}} \]
Taking the square root:
\[ \text{LHS} = \frac{\cos A}{1 - \sin A} = \text{RHS} \]
Hence proved.
In simple words: Multiply inside the square root by the conjugate of the denominator to form a perfect square. This allows you to evaluate the square root easily.

Exam Tip: Rationalizing or converting the expression inside a square root to perfect squares is the standard way to clear the radical symbol in proofs.

Question 36. Prove that \(\sin (90 - \Theta) \cos (90 - \Theta) = \frac{\tan\Theta}{1 + \tan^2\Theta}\)
Answer:
Let's simplify the Left-Hand Side (LHS):
Using complementary angle identities:
- \(\sin(90^\circ - \Theta) = \cos\Theta\)
- \(\cos(90^\circ - \Theta) = \sin\Theta\)
So, \(\text{LHS} = \cos\Theta\sin\Theta\).

Now, let's simplify the Right-Hand Side (RHS):
\[ \text{RHS} = \frac{\tan\Theta}{1 + \tan^2\Theta} \]
Using the identity \(1 + \tan^2\Theta = \sec^2\Theta\):
\[ \text{RHS} = \frac{\tan\Theta}{\sec^2\Theta} \]
Convert to sine and cosine:
\[ \text{RHS} = \frac{\frac{\sin\Theta}{\cos\Theta}}{\frac{1}{\cos^2\Theta}} = \frac{\sin\Theta}{\cos\Theta} \cdot \cos^2\Theta = \sin\Theta\cos\Theta \]
Since \(\text{LHS} = \text{RHS}\), the identity is proved.
In simple words: Convert the complementary ratios on the left side to get sine and cosine. Convert the right side terms using the secant identity to show both sides are equal.

Exam Tip: If one side of a proof is difficult to simplify directly to the other, try simplifying both sides to a common intermediate expression.

Question 37. If \(x = a \sec\Theta + b \tan\Theta\) and \(y = a \tan\Theta + b \sec\Theta\) prove that \(x^2 - y^2 = a^2 - b^2\)
Answer:
Given equations:
1. \(x = a\sec\Theta + b\tan\Theta\)
2. \(y = a\tan\Theta + b\sec\Theta\)
Let's expand the terms in the target equation, \(x^2 - y^2\):
\[ x^2 = (a\sec\Theta + b\tan\Theta)^2 = a^2\sec^2\Theta + b^2\tan^2\Theta + 2ab\sec\Theta\tan\Theta \]
\[ y^2 = (a\tan\Theta + b\sec\Theta)^2 = a^2\tan^2\Theta + b^2\sec^2\Theta + 2ab\sec\Theta\tan\Theta \]
Now, subtract \(y^2\) from \(x^2\):
\[ x^2 - y^2 = (a^2\sec^2\Theta + b^2\tan^2\Theta + 2ab\sec\Theta\tan\Theta) - (a^2\tan^2\Theta + b^2\sec^2\Theta + 2ab\sec\Theta\tan\Theta) \]
The cross-product terms \(2ab\sec\Theta\tan\Theta\) cancel out:
\[ x^2 - y^2 = a^2\sec^2\Theta - a^2\tan^2\Theta + b^2\tan^2\Theta - b^2\sec^2\Theta \]
Factor out \(a^2\) and \(-b^2\):
\[ x^2 - y^2 = a^2(\sec^2\Theta - \tan^2\Theta) - b^2(\sec^2\Theta - \tan^2\Theta) \]
Using the identity \(\sec^2\Theta - \tan^2\Theta = 1\):
\[ x^2 - y^2 = a^2(1) - b^2(1) = a^2 - b^2 = \text{RHS} \]
Hence proved.
In simple words: Square both expressions algebraically. When you subtract them, the mixed terms cancel out, leaving standard secant-tangent identity terms that simplify to 1.

Exam Tip: Be careful with signs when subtracting bracketed algebraic expansions to avoid sign errors on grouped terms.

Question 38. Show that \(\frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = \sin A + \cos A\)
Answer:
Let's simplify the Left-Hand Side (LHS):
\[ \text{LHS} = \frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} \]
Substitute \(\tan A = \frac{\sin A}{\cos A}\) and \(\cot A = \frac{\cos A}{\sin A}\):
\[ \text{LHS} = \frac{\cos A}{1 - \frac{\sin A}{\cos A}} + \frac{\sin A}{1 - \frac{\cos A}{\sin A}} \]
Simplify the denominators:
\[ \text{LHS} = \frac{\cos A}{\frac{\cos A - \sin A}{\cos A}} + \frac{\sin A}{\frac{\sin A - \cos A}{\sin A}} \]
\[ \text{LHS} = \frac{\cos^2 A}{\cos A - \sin A} + \frac{\sin^2 A}{\sin A - \cos A} \]
Make the denominators common by rewriting \(\sin A - \cos A\) as \(-(\cos A - \sin A)\):
\[ \text{LHS} = \frac{\cos^2 A}{\cos A - \sin A} - \frac{\sin^2 A}{\cos A - \sin A} \]
Combine the fractions:
\[ \text{LHS} = \frac{\cos^2 A - \sin^2 A}{\cos A - \sin A} \]
Factor the numerator using the difference of squares identity \(a^2 - b^2 = (a-b)(a+b)\):
\[ \text{LHS} = \frac{(\cos A - \sin A)(\cos A + \sin A)}{\cos A - \sin A} \]
Cancel the common binomial term \(\cos A - \sin A\):
\[ \text{LHS} = \cos A + \sin A = \text{RHS} \]
Hence proved.
In simple words: Express tangent and cotangent in terms of sine and cosine. Finding a common denominator allows you to combine the terms and factor the numerator to cancel the denominator.

Exam Tip: Converting everything to sine and cosine is a highly reliable fallback strategy for almost all trigonometric identity proofs.

Question 39. Prove that \(\sec^2\Theta + \csc^2\Theta = \sec^2\Theta \cdot \csc^2\Theta\)
Answer:
Let's simplify the Left-Hand Side (LHS):
\[ \text{LHS} = \sec^2\Theta + \csc^2\Theta \]
Express in terms of sine and cosine:
\[ \text{LHS} = \frac{1}{\cos^2\Theta} + \frac{1}{\sin^2\Theta} \]
Find a common denominator:
\[ \text{LHS} = \frac{\sin^2\Theta + \cos^2\Theta}{\cos^2\Theta \cdot \sin^2\Theta} \]
Using the identity \(\sin^2\Theta + \cos^2\Theta = 1\):
\[ \text{LHS} = \frac{1}{\cos^2\Theta \cdot \sin^2\Theta} \]
Separate the fraction into products:
\[ \text{LHS} = \left(\frac{1}{\cos^2\Theta}\right) \left(\frac{1}{\sin^2\Theta}\right) = \sec^2\Theta \cdot \csc^2\Theta = \text{RHS} \]
Hence proved.
In simple words: Convert the terms to sine and cosine. Combining them over a single denominator uses the identity \(\sin^2\Theta + \cos^2\Theta = 1\) to yield the product form.

Exam Tip: Remember that adding the reciprocals of squared sine and cosine terms naturally transforms their sum into a product.

Question 40. Prove that \(\frac{\cot\Theta}{1 + \tan\Theta} = \frac{\cot\Theta - 1}{2 - \sec^2\Theta}\)
Answer:
Let's simplify the Left-Hand Side (LHS):
\[ \text{LHS} = \frac{\cot\Theta}{1 + \tan\Theta} = \frac{\frac{1}{\tan\Theta}}{1 + \tan\Theta} = \frac{1}{\tan\Theta(1 + \tan\Theta)} \]

Now, let's simplify the Right-Hand Side (RHS):
\[ \text{RHS} = \frac{\cot\Theta - 1}{2 - \sec^2\Theta} \]
Substitute \(\cot\Theta = \frac{1}{\tan\Theta}\) and \(\sec^2\Theta = 1 + \tan^2\Theta\):
\[ \text{RHS} = \frac{\frac{1}{\tan\Theta} - 1}{2 - (1 + \tan^2\Theta)} \]
Simplify the numerator and denominator:
\[ \text{RHS} = \frac{\frac{1 - \tan\Theta}{\tan\Theta}}{1 - \tan^2\Theta} \]
\[ \text{RHS} = \frac{1 - \tan\Theta}{\tan\Theta(1 - \tan^2\Theta)} \]
Using \(1 - \tan^2\Theta = (1 - \tan\Theta)(1 + \tan\Theta)\):
\[ \text{RHS} = \frac{1 - \tan\Theta}{\tan\Theta(1 - \tan\Theta)(1 + \tan\Theta)} \]
Cancel the common factor \((1 - \tan\Theta)\):
\[ \text{RHS} = \frac{1}{\tan\Theta(1 + \tan\Theta)} \]
Since \(\text{LHS} = \text{RHS}\), the identity is proved.
In simple words: Convert the cotangent and secant terms to tangent. This simplifies both sides to the exact same algebraic rational expression in terms of tangent.

Exam Tip: Converting a mixed expression involving secant, tangent, and cotangent entirely into tangent can be a great way to simplify the expression.

Question 41. Prove that \(\frac{1 - \sin\Theta}{1 + \sin\Theta} = (\sec\Theta - \tan\Theta )^2\)
Answer:
Let's evaluate the Right-Hand Side (RHS):
\[ \text{RHS} = (\sec\Theta - \tan\Theta)^2 \]
Substitute \(\sec\Theta = \frac{1}{\cos\Theta}\) and \(\tan\Theta = \frac{\sin\Theta}{\cos\Theta}\):
\[ \text{RHS} = \left(\frac{1}{\cos\Theta} - \frac{\sin\Theta}{\cos\Theta}\right)^2 \]
Combine the terms under a common denominator:
\[ \text{RHS} = \left(\frac{1 - \sin\Theta}{\cos\Theta}\right)^2 \]
\[ \text{RHS} = \frac{(1 - \sin\Theta)^2}{\cos^2\Theta} \]
Substitute \(\cos^2\Theta = 1 - \sin^2\Theta\):
\[ \text{RHS} = \frac{(1 - \sin\Theta)^2}{1 - \sin^2\Theta} \]
Factor the denominator using the difference of squares: \(1 - \sin^2\Theta = (1 - \sin\Theta)(1 + \sin\Theta)\):
\[ \text{RHS} = \frac{(1 - \sin\Theta)^2}{(1 - \sin\Theta)(1 + \sin\Theta)} \]
Cancel out the common term \((1 - \sin\Theta)\):
\[ \text{RHS} = \frac{1 - \sin\Theta}{1 + \sin\Theta} = \text{LHS} \]
Hence proved.
In simple words: Convert the right-hand side to sine and cosine, group the fraction, and substitute \(\cos^2\Theta = 1 - \sin^2\Theta\) to cancel common factors and reach the left-hand side.

Exam Tip: Working backwards from the more complex side (often the squared term) to the simpler ratio side is a great strategy in proofs.

Question 42. Prove that: \(\tan^2 A - \tan^2 B = \frac{\sin^2 A - \sin^2 B}{\cos^2 A \cdot \cos^2 B}\)
Answer:
Let's evaluate the Left-Hand Side (LHS):
\[ \text{LHS} = \tan^2 A - \tan^2 B \]
Convert tangent to sine and cosine:
\[ \text{LHS} = \frac{\sin^2 A}{\cos^2 A} - \frac{\sin^2 B}{\cos^2 B} \]
Find a common denominator:
\[ \text{LHS} = \frac{\sin^2 A \cos^2 B - \sin^2 B \cos^2 A}{\cos^2 A \cdot \cos^2 B} \]
Using the identity \(\cos^2\theta = 1 - \sin^2\theta\) to convert the numerator terms to sines:
\[ \text{LHS} = \frac{\sin^2 A (1 - \sin^2 B) - \sin^2 B (1 - \sin^2 A)}{\cos^2 A \cdot \cos^2 B} \]
Expand the numerator:
\[ \text{LHS} = \frac{\sin^2 A - \sin^2 A \sin^2 B - \sin^2 B + \sin^2 B \sin^2 A}{\cos^2 A \cdot \cos^2 B} \]
The term \(\sin^2 A \sin^2 B\) cancels out:
\[ \text{LHS} = \frac{\sin^2 A - \sin^2 B}{\cos^2 A \cdot \cos^2 B} = \text{RHS} \]
Hence proved.
In simple words: Convert the tangent squared terms to fractions of sine and cosine. Use cross-multiplication to combine them, and rewrite the cosines in the numerator to simplify and cancel terms.

Exam Tip: Be careful with the distributive property when expanding the subtraction in the numerator to keep your signs correct.

Question 43. Prove that : \((\sin\Theta + \csc\Theta)^2 + (\cos\Theta + \sec\Theta)^2 = 7 + \tan^2\Theta + \cot^2\Theta\)
Answer:
Let's expand the Left-Hand Side (LHS):
\[ \text{LHS} = (\sin\Theta + \csc\Theta)^2 + (\cos\Theta + \sec\Theta)^2 \]
\[ \text{LHS} = (\sin^2\Theta + \csc^2\Theta + 2\sin\Theta\csc\Theta) + (\cos^2\Theta + \sec^2\Theta + 2\cos\Theta\sec\Theta) \]
Since \(\sin\Theta \cdot \csc\Theta = 1\) and \(\cos\Theta \cdot \sec\Theta = 1\):
\[ \text{LHS} = \sin^2\Theta + \csc^2\Theta + 2(1) + \cos^2\Theta + \sec^2\Theta + 2(1) \]
\[ \text{LHS} = \sin^2\Theta + \cos^2\Theta + \csc^2\Theta + \sec^2\Theta + 4 \]
Since \(\sin^2\Theta + \cos^2\Theta = 1\):
\[ \text{LHS} = 1 + \csc^2\Theta + \sec^2\Theta + 4 = 5 + \csc^2\Theta + \sec^2\Theta \]
Using the Pythagorean identities \(\csc^2\Theta = 1 + \cot^2\Theta\) and \(\sec^2\Theta = 1 + \tan^2\Theta\):
\[ \text{LHS} = 5 + (1 + \cot^2\Theta) + (1 + \tan^2\Theta) \]
\[ \text{LHS} = 7 + \tan^2\Theta + \cot^2\Theta = \text{RHS} \]
Hence proved.
In simple words: Expand the squared terms algebraically. Use the reciprocal identities to simplify the cross-products to constants, then convert the remaining squares using standard Pythagorean relations to get the final constant of 7.

Exam Tip: Expanding squares is often the easiest entry point for trigonometric proofs when you have binomial terms.

Question 44. Prove that \((\csc\Theta - \cot\Theta)^2 = \frac{1 - \cos\Theta}{1 + \cos\Theta}\)
Answer:
Let's simplify the Left-Hand Side (LHS):
\[ \text{LHS} = (\csc\Theta - \cot\Theta)^2 \]
Convert to sine and cosine:
\[ \text{LHS} = \left(\frac{1}{\sin\Theta} - \frac{\cos\Theta}{\sin\Theta}\right)^2 \]
Combine under a common denominator:
\[ \text{LHS} = \left(\frac{1 - \cos\Theta}{\sin\Theta}\right)^2 = \frac{(1 - \cos\Theta)^2}{\sin^2\Theta} \]
Using the identity \(\sin^2\Theta = 1 - \cos^2\Theta\):
\[ \text{LHS} = \frac{(1 - \cos\Theta)^2}{1 - \cos^2\Theta} \]
Factor the denominator as \((1 - \cos\Theta)(1 + \cos\Theta)\):
\[ \text{LHS} = \frac{(1 - \cos\Theta)^2}{(1 - \cos\Theta)(1 + \cos\Theta)} \]
Cancel the common factor \((1 - \cos\Theta)\):
\[ \text{LHS} = \frac{1 - \cos\Theta}{1 + \cos\Theta} = \text{RHS} \]
Hence proved.
In simple words: Convert the terms to sine and cosine. Simplify the resulting fraction, convert sine squared to cosine, and cancel the common factor.

Exam Tip: Factoring the denominator using the difference of squares identity is a key algebraic step to simplify rational fractions in trigonometry.

Question 45. Prove that \(\frac{1}{\sec\Theta - \tan\Theta} - \frac{1}{\cos\Theta} = \frac{1}{\cos\Theta} - \frac{1}{\sec\Theta + \tan\Theta}\)
Answer:
The given equation can be rearranged to prove a simpler, equivalent relation:
\[ \frac{1}{\sec\Theta - \tan\Theta} + \frac{1}{\sec\Theta + \tan\Theta} = \frac{1}{\cos\Theta} + \frac{1}{\cos\Theta} = \frac{2}{\cos\Theta} = 2\sec\Theta \]
Let's evaluate the Left-Hand Side (LHS) of this rearranged equation:
\[ \text{LHS} = \frac{1}{\sec\Theta - \tan\Theta} + \frac{1}{\sec\Theta + \tan\Theta} \]
Find a common denominator:
\[ \text{LHS} = \frac{(\sec\Theta + \tan\Theta) + (\sec\Theta - \tan\Theta)}{(\sec\Theta - \tan\Theta)(\sec\Theta + \tan\Theta)} \]
Simplify the numerator and denominator:
\[ \text{LHS} = \frac{2\sec\Theta}{\sec^2\Theta - \tan^2\Theta} \]
Using the identity \(\sec^2\Theta - \tan^2\Theta = 1\):
\[ \text{LHS} = \frac{2\sec\Theta}{1} = 2\sec\Theta = \frac{2}{\cos\Theta} = \text{RHS} \]
Hence proved.
In simple words: Rearrange the terms by grouping similar expressions on the same side. This creates a sum of conjugate fractions which simplifies easily using standard identities.

Exam Tip: Rearranging the target equation before starting the proof is a completely valid and extremely helpful mathematical approach to simplify the algebraic steps.

Question 46. Prove that \(\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A\)
Answer:
Let's simplify the Left-Hand Side (LHS):
\[ \text{LHS} = \sqrt{\frac{1 + \sin A}{1 - \sin A}} \]
Multiply the numerator and the denominator inside the square root by \((1 + \sin A)\):
\[ \text{LHS} = \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}} \]
\[ \text{LHS} = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}} \]
Using the identity \(1 - \sin^2 A = \cos^2 A\):
\[ \text{LHS} = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} \]
Taking the square root:
\[ \text{LHS} = \frac{1 + \sin A}{\cos A} \]
Split the fraction into two terms:
\[ \text{LHS} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A = \text{RHS} \]
Hence proved.
In simple words: Multiply the top and bottom of the fraction inside the square root by \((1+\sin A)\). This rationalizes the denominator to cosine squared, allowing you to remove the radical sign.

Exam Tip: Multiplying by the conjugate of the denominator is a standard algebraic method to eliminate square roots in trigonometric proofs.

Question 47. Prove that \(\sec^4 \Theta - \tan^4 \Theta = 1 + 2 \tan^2 \Theta\)
Answer:
Let's evaluate the Left-Hand Side (LHS):
\[ \text{LHS} = \sec^4\Theta - \tan^4\Theta \]
This is in the form of a difference of squares \(a^2 - b^2\) where \(a = \sec^2\Theta\) and \(b = \tan^2\Theta\):
\[ \text{LHS} = (\sec^2\Theta - \tan^2\Theta)(\sec^2\Theta + \tan^2\Theta) \]
Since \(\sec^2\Theta - \tan^2\Theta = 1\):
\[ \text{LHS} = 1 \cdot (\sec^2\Theta + \tan^2\Theta) = \sec^2\Theta + \tan^2\Theta \]
Now, substitute \(\sec^2\Theta = 1 + \tan^2\Theta\):
\[ \text{LHS} = (1 + \tan^2\Theta) + \tan^2\Theta = 1 + 2\tan^2\Theta = \text{RHS} \]
Hence proved.
In simple words: Apply the difference of squares algebraic identity first. This reduces the fourth-degree terms to standard squared terms, allowing you to use the basic secant-tangent identity.

Exam Tip: Factoring polynomials with trigonometric terms works exactly like factoring normal algebraic expressions, so use standard formulas like difference of squares.

Question 48. Show that \(\frac{\sin\Theta - 2 \sin^3\Theta}{2 \cos^3\Theta - \cos\Theta} = \tan\Theta\)
Answer:
Let's simplify the Left-Hand Side (LHS):
\[ \text{LHS} = \frac{\sin\Theta - 2\sin^3\Theta}{2\cos^3\Theta - \cos\Theta} \]
Factor out \(\sin\Theta\) from the numerator and \(\cos\Theta\) from the denominator:
\[ \text{LHS} = \frac{\sin\Theta (1 - 2\sin^2\Theta)}{\cos\Theta (2\cos^2\Theta - 1)} \]
We know the identities for \(\cos 2\Theta\):
- \(1 - 2\sin^2\Theta = \cos 2\Theta\)
- \(2\cos^2\Theta - 1 = \cos 2\Theta\)
Substituting these identities:
\[ \text{LHS} = \frac{\sin\Theta \cdot \cos 2\Theta}{\cos\Theta \cdot \cos 2\Theta} \]
Cancel the common term \(\cos 2\Theta\):
\[ \text{LHS} = \frac{\sin\Theta}{\cos\Theta} = \tan\Theta = \text{RHS} \]
Hence proved.
In simple words: Factor out sine and cosine from the top and bottom. The remaining terms are different forms of the same identity, so they cancel out, leaving just tangent.

Exam Tip: Even if you haven't studied \(\cos 2\theta\) formulas yet, you can prove they are equal by substituting \(1 - \sin^2\theta\) for \(\cos^2\theta\) in the denominator to match the numerator.

Question 49. If \(\sec\Theta + \tan\Theta = p\), prove that \(\sin\Theta = \frac{p^2 - 1}{p^2 + 1}\)
Answer:
We are given:
\[ \sec\Theta + \tan\Theta = p \quad \text{--- (Equation 1)} \]
We know the identity \(\sec^2\Theta - \tan^2\Theta = 1\), which can be written as \(( \sec\Theta - \tan\Theta )( \sec\Theta + \tan\Theta ) = 1\).
Substituting the value of Equation 1:
\[ \sec\Theta - \tan\Theta = \frac{1}{p} \quad \text{--- (Equation 2)} \]
Adding Equation 1 and Equation 2:
\[ 2\sec\Theta = p + \frac{1}{p} = \frac{p^2 + 1}{p} \implies \sec\Theta = \frac{p^2 + 1}{2p} \]
Subtracting Equation 2 from Equation 1:
\[ 2\tan\Theta = p - \frac{1}{p} = \frac{p^2 - 1}{p} \implies \tan\Theta = \frac{p^2 - 1}{2p} \]
Now, we know that \(\sin\Theta = \frac{\tan\Theta}{\sec\Theta}\). Substituting our values:
\[ \sin\Theta = \frac{\frac{p^2 - 1}{2p}}{\frac{p^2 + 1}{2p}} = \frac{p^2 - 1}{p^2 + 1} = \text{RHS} \]
Hence proved.
In simple words: Establish two equations using the secant-tangent identity. Solve them for secant and tangent individually, then divide the two values to obtain the expression for sine.

Exam Tip: This is a standard proof. Remember that if \(\sec\theta + \tan\theta = p\), then its conjugate is always the reciprocal \(1/p\).

Question 50. Prove that \(\frac{\tan\Theta + \sin\Theta}{\tan\Theta - \sin\Theta} = \frac{\sec\Theta + 1}{\sec\Theta - 1}\)
Answer:
Let's simplify the Left-Hand Side (LHS):
\[ \text{LHS} = \frac{\tan\Theta + \sin\Theta}{\tan\Theta - \sin\Theta} \]
Convert tangent to sine and cosine:
\[ \text{LHS} = \frac{\frac{\sin\Theta}{\cos\Theta} + \sin\Theta}{\frac{\sin\Theta}{\cos\Theta} - \sin\Theta} \]
Factor out \(\sin\Theta\) from both the numerator and the denominator:
\[ \text{LHS} = \frac{\sin\Theta \left(\frac{1}{\cos\Theta} + 1\right)}{\sin\Theta \left(\frac{1}{\cos\Theta} - 1\right)} \]
Cancel the common term \(\sin\Theta\):
\[ \text{LHS} = \frac{\frac{1}{\cos\Theta} + 1}{\frac{1}{\cos\Theta} - 1} \]
Substitute \(\frac{1}{\cos\Theta} = \sec\Theta\):
\[ \text{LHS} = \frac{\sec\Theta + 1}{\sec\Theta - 1} = \text{RHS} \]
Hence proved.
In simple words: Convert the tangent term to sine over cosine. Factoring out sine from both numerator and denominator allows you to simplify the remaining terms directly to secant.

Exam Tip: Factoring out common trigonometric functions (like sine) is a quick way to clean up rational fractions and simplify them instantly.