CBSE Class 10 Mathematics Some Applications of Trigonometry Worksheet Set 07

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)

Section A: (1 Mark)

Question 1. If sin (A + B) = 1 and cos (A – B) = \(\frac{\sqrt{3}}{2}\), then find A and B.
Answer: We are given that \(\sin(A+B) = 1\). Since we know that \(\sin 90^\circ = 1\), this gives us our first relation: \(A + B = 90^\circ\) — (1) Next, we have \(\cos(A-B) = \frac{\sqrt{3}}{2}\). Since \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), this gives us our second relation: \(A - B = 30^\circ\) — (2) To determine the values of our variables, we add equations (1) and (2) together: \((A + B) + (A - B) = 90^\circ + 30^\circ\)
\(\implies 2A = 120^\circ\)
\(\implies A = 60^\circ\) Now, substituting this value of \(A\) into equation (1) yields: \(60^\circ + B = 90^\circ\)
\(\implies B = 30^\circ\) Hence, the angles are \(A = 60^\circ\) and \(B = 30^\circ\).
In simple words: Match the trigonometric values with their standard angles to get two linear equations. Solving them simultaneously gives \(A = 60^\circ\) and \(B = 30^\circ\).

Exam Tip: Be sure to write the equations clearly and verify that your final angle values are positive and acute.

Question 2. Express sin 67⁰ + cos75⁰ in terms of trigonometric ratios of angles between 0⁰ and 45⁰.
Answer: We can rewrite the angles in terms of complementary angles. We know that: \(\sin \theta = \cos(90^\circ - \theta)\) \(\cos \theta = \sin(90^\circ - \theta)\) By applying these identities to our given expression: \(\sin 67^\circ = \cos(90^\circ - 67^\circ) = \cos 23^\circ\) \(\cos 75^\circ = \sin(90^\circ - 75^\circ) = \sin 15^\circ\) Thus, the expression \(\sin 67^\circ + \cos 75^\circ\) is represented as: \(\cos 23^\circ + \sin 15^\circ\) Both \(23^\circ\) and \(15^\circ\) are within the specified range of \(0^\circ\) to \(45^\circ\).
In simple words: Swap sine and cosine with their complementary functions to bring the angles down to values less than \(45^\circ\).

Exam Tip: Always show the subtraction from \(90^\circ\) to earn full method marks.

Question 3. If tan A = cot B, prove that A + B = 90⁰
Answer: We begin with the given equation: \(\tan A = \cot B\) We can make use of the complementary identity: \(\cot B = \tan(90^\circ - B)\). Substituting this on the right-hand side of our equation, we get: \(\tan A = \tan(90^\circ - B)\) Equating the angles on both sides, we find: \(A = 90^\circ - B\)
\(\implies A + B = 90^\circ\) The relationship is proved.
In simple words: Since \(\tan A\) and \(\cot B\) are equal, they must be complementary angles. Converting cotangent to tangent makes this clear.

Exam Tip: State the co-function identity clearly in brackets next to your step to make your proof rigorous.

Section B: (2 Marks)

Question 4. If tan² (3A + 15)⁰ - 1 = 0, then find the value of A that satisfy this condition.
Answer: We are given the equation: \(\tan^2(3A + 15)^\circ - 1 = 0\) Adding 1 to both sides gives: \(\tan^2(3A + 15)^\circ = 1\) Taking the positive square root for acute angles, we get: \(\tan(3A + 15)^\circ = 1\) Since we know that \(\tan 45^\circ = 1\), we can equate the angles: \(3A + 15 = 45\) Subtracting 15 from both sides yields: \(3A = 30\)
\(\implies A = 10^\circ\) Thus, the value of \(A\) is \(10^\circ\).
In simple words: Move the constant to the other side and take the square root. Equate the angle to \(45^\circ\) and solve the simple linear equation.

Exam Tip: Be sure to keep track of the degree signs and check your arithmetic at the end.

Question 5. Prove that ( √3 + 1) (3 – cot 30⁰) = tan³ 60⁰ - 2 sin 60⁰
Answer: Let us evaluate the Left Hand Side (LHS) of the expression: \(\text{LHS} = (\sqrt{3} + 1)(3 - \cot 30^\circ)\) Since we know that \(\cot 30^\circ = \sqrt{3}\), we substitute this in: \(\text{LHS} = (\sqrt{3} + 1)(3 - \sqrt{3})\) Multiplying out the terms: \(\text{LHS} = 3\sqrt{3} - \sqrt{3}\cdot\sqrt{3} + 3 - \sqrt{3}\) \(\text{LHS} = 3\sqrt{3} - 3 + 3 - \sqrt{3} = 2\sqrt{3}\) Now, let us evaluate the Right Hand Side (RHS): \(\text{RHS} = \tan^3 60^\circ - 2\sin 60^\circ\) We know that \(\tan 60^\circ = \sqrt{3}\) and \(\sin 60^\circ = \frac{\sqrt{3}}{2}\). Substituting these values: \(\text{RHS} = (\sqrt{3})^3 - 2\left(\frac{\sqrt{3}}{2}\right)\) \(\text{RHS} = 3\sqrt{3} - \sqrt{3} = 2\sqrt{3}\) Since both LHS and RHS are equal to \(2\sqrt{3}\), the identity holds true.
In simple words: Replace all trigonometric terms with their standard values, simplify both sides of the equation independently, and verify that they are equal.

Exam Tip: Remembering the standard values of \(30^\circ\) and \(60^\circ\) correctly is vital to avoid algebraic errors.

Question 6. If sinϴ + sin²ϴ = 1, check the validity of the expression: cos²ϴ + cos⁴ϴ = 1.
Answer: We are given the relation: \(\sin\theta + \sin^2\theta = 1\) We can rewrite this as: \(\sin\theta = 1 - \sin^2\theta\) Using the standard Pythagorean identity \(\cos^2\theta = 1 - \sin^2\theta\), we get: \(\sin\theta = \cos^2\theta\) — (1) Squaring both sides of equation (1): \(\sin^2\theta = \cos^4\theta\) — (2) Let us substitute equation (2) into the given expression to check its validity: \(\text{LHS} = \cos^2\theta + \cos^4\theta\) \(\text{LHS} = \cos^2\theta + \sin^2\theta\) Since \(\cos^2\theta + \sin^2\theta = 1\), we have: \(\text{LHS} = 1\) This matches the right-hand side of the expression. Thus, the expression \(\cos^2\theta + \cos^4\theta = 1\) is valid.
In simple words: Rewrite the given equation to express \(\sin\theta\) in terms of \(\cos^2\theta\). By substituting this back, the expression turns into the standard identity \(\sin^2\theta + \cos^2\theta = 1\).

Exam Tip: Substituting higher-order powers of cosine with lower-order powers of sine makes identity checking much simpler.

Question 7. If sin ϴ = 𝑎/b, then find secϴ + tanϴ in terms of a and b.
Answer: We are given: \(\sin\theta = \frac{a}{b}\) In a right-angled triangle, let the opposite side be \(a\) and the hypotenuse be \(b\). By Pythagoras' theorem, the adjacent side is: \(\text{Adjacent Side} = \sqrt{b^2 - a^2}\) Now, we find \(\sec\theta\) and \(\tan\theta\) using their definitions: \(\sec\theta = \frac{\text{Hypotenuse}}{\text{Adjacent Side}} = \frac{b}{\sqrt{b^2 - a^2}}\) \(\tan\theta = \frac{\text{Opposite}}{\text{Adjacent Side}} = \frac{a}{\sqrt{b^2 - a^2}}\) Adding these two ratios together, we get: \(\sec\theta + \tan\theta = \frac{b + a}{\sqrt{b^2 - a^2}}\) Using the algebraic identity \(b^2 - a^2 = (b - a)(b + a)\) in the denominator: \(\sec\theta + \tan\theta = \frac{b + a}{\sqrt{(b - a)(b + a)}}\) We can rewrite the numerator as \(\sqrt{b + a} \cdot \sqrt{b + a}\): \(\sec\theta + \tan\theta = \frac{\sqrt{b + a} \cdot \sqrt{b + a}}{\sqrt{b - a} \cdot \sqrt{b + a}}\) Cancelling the common term \(\sqrt{b + a}\): \(\sec\theta + \tan\theta = \sqrt{\frac{b + a}{b - a}}\)
In simple words: Find the adjacent side of the triangle using the Pythagorean theorem, write down the ratios for \(\sec\theta\) and \(\tan\theta\), then simplify the resulting algebraic fraction.

Exam Tip: Simplifying the root expression in the final step is essential to match the standard format and score full marks.

Section C: (3 Marks)

Question 8. If sin ϴ = cos ϴ, find the value of 2 tan²ϴ + sin²ϴ - 1
Answer: We are given that: \(\sin\theta = \cos\theta\) Dividing both sides by \(\cos\theta\) gives: \(\tan\theta = 1\) Since \(\tan 45^\circ = 1\), we have \(\theta = 45^\circ\). We can now substitute \(\theta = 45^\circ\) into our expression: \(\text{Value} = 2\tan^2(45^\circ) + \sin^2(45^\circ) - 1\) Using standard values \(\tan 45^\circ = 1\) and \(\sin 45^\circ = \frac{1}{\sqrt{2}}\): \(\text{Value} = 2(1)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 - 1\) \(\text{Value} = 2 + \frac{1}{2} - 1 = 1 + \frac{1}{2} = \frac{3}{2}\) Thus, the value of the expression is \(\frac{3}{2}\).
In simple words: The given condition tells us that \(\theta = 45^\circ\). Substituting this angle into the expression lets us solve it using standard values.

Exam Tip: Remember that \(\sin\theta = \cos\theta\) always implies \(\theta = 45^\circ\) in the first quadrant.

Question 9. Find the value of sin 30⁰, sin 45⁰ and sin 60⁰ geometrically.
Answer: Let us derive these standard values geometrically.
• For \(30^\circ\) and \(60^\circ\):
Consider an equilateral triangle \(ABC\) of side length \(2a\). In an equilateral triangle, each angle is \(60^\circ\). Draw a perpendicular altitude \(AD\) from \(A\) to \(BC\). This bisects the vertical angle and the base, giving \(BD = a\) and \(\angle BAD = 30^\circ\). In right-angled triangle \(ABD\), using Pythagoras' theorem: \(AD = \sqrt{AB^2 - BD^2} = \sqrt{(2a)^2 - a^2} = a\sqrt{3}\)
(i) \(\sin 30^\circ = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BD}{AB} = \frac{a}{2a} = \frac{1}{2}\)
(ii) \(\sin 60^\circ = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AD}{AB} = \frac{a\sqrt{3}}{2a} = \frac{\sqrt{3}}{2}\)

• For \(45^\circ\):
Consider an isosceles right-angled triangle \(XYZ\) with right angle at \(Y\) and equal sides \(XY = YZ = a\). Since it is an isosceles right triangle, \(\angle X = \angle Z = 45^\circ\). Using Pythagoras' theorem to find the hypotenuse \(XZ\): \(XZ = \sqrt{XY^2 + YZ^2} = \sqrt{a^2 + a^2} = a\sqrt{2}\) Using the definition of sine for angle \(X\): \(\sin 45^\circ = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{YZ}{XZ} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}}\)

In simple words: Construct an equilateral triangle to prove the values of \(30^\circ\) and \(60^\circ\) ratios, and use an isosceles right-angled triangle to derive the \(45^\circ\) ratio.

Exam Tip: Be sure to draw clean, labeled diagrams of both triangles to secure full presentation marks.

Question 10. Find the value of the following : \[\frac{\cos 50^\circ}{2\sin 40^\circ} + \frac{4(\csc^2 59^\circ - \tan^2 31^\circ)}{3\tan^2 45^\circ} - \frac{2}{3}\tan 12^\circ \tan 78^\circ \sin 90^\circ\]
Answer: Let us simplify each term step by step.
• First Term: \(\frac{\cos 50^\circ}{2\sin 40^\circ}\) Since \(\cos(90^\circ - \theta) = \sin\theta\), we can write: \(\cos 50^\circ = \cos(90^\circ - 40^\circ) = \sin 40^\circ\) Substituting this value: \(\frac{\sin 40^\circ}{2\sin 40^\circ} = \frac{1}{2}\)
• Second Term: \(\frac{4(\csc^2 59^\circ - \tan^2 31^\circ)}{3\tan^2 45^\circ}\) Using co-function identities, we know that \(\tan 31^\circ = \cot(90^\circ - 31^\circ) = \cot 59^\circ\). The term \(\csc^2 59^\circ - \cot^2 59^\circ\) simplifies to 1 using the standard identity \(\csc^2\theta - \cot^2\theta = 1\). Also, the standard value of \(\tan 45^\circ = 1\), so the term becomes: \(\frac{4(1)}{3(1)^2} = \frac{4}{3}\)
• Third Term: \(\frac{2}{3}\tan 12^\circ \tan 78^\circ \sin 90^\circ\) Using complementary relations, \(\tan 78^\circ = \cot(90^\circ - 78^\circ) = \cot 12^\circ\). Since \(\tan 12^\circ \cdot \cot 12^\circ = 1\) and \(\sin 90^\circ = 1\), this term evaluates to: \(\frac{2}{3}(1)(1) = \frac{2}{3}\)
Now, combining all three simplified terms: \(\text{Value} = \frac{1}{2} + \frac{4}{3} - \frac{2}{3}\) \(\text{Value} = \frac{1}{2} + \frac{2}{3} = \frac{3 + 4}{6} = \frac{7}{6}\) Thus, the value is \(\frac{7}{6}\).
In simple words: Convert the complementary angles in each fraction to matching trigonometric functions, apply basic identities, and sum the simple fractions to get \(\frac{7}{6}\).

Exam Tip: Be sure to write the formula for complementary angles (e.g., \(\tan(90^\circ - \theta) = \cot\theta\)) next to the step where you apply it to ensure you get full process credit.

Question 11. If x\(\sin^3\theta\) + y\(\cos^3\theta\) = \(\sin\theta \cos\theta\) and x sin ϴ = y cos ϴ, prove that x² + y² = 1.
Answer: We are given the equations: \(x\sin^3\theta + y\cos^3\theta = \sin\theta \cos\theta\) — (1) \(x\sin\theta = y\cos\theta\) — (2) We can rewrite equation (1) as: \((x\sin\theta)\sin^2\theta + (y\cos\theta)\cos^2\theta = \sin\theta \cos\theta\) Now, substituting the value of \(y\cos\theta = x\sin\theta\) from equation (2) into this equation: \((x\sin\theta)\sin^2\theta + (x\sin\theta)\cos^2\theta = \sin\theta \cos\theta\) Factoring out \(x\sin\theta\) from the left-hand side: \(x\sin\theta(\sin^2\theta + \cos^2\theta) = \sin\theta \cos\theta\) Since we know that \(\sin^2\theta + \cos^2\theta = 1\), this simplifies to: \(x\sin\theta = \sin\theta \cos\theta\) Dividing both sides by \(\sin\theta\): \(x = \cos\theta\) — (3) By substituting equation (3) into equation (2): \(\cos\theta \sin\theta = y\cos\theta\) Dividing both sides by \(\cos\theta\): \(y = \sin\theta\) — (4) Now, squaring and adding equations (3) and (4): \(x^2 + y^2 = \cos^2\theta + \sin^2\theta\) Applying the identity \(\cos^2\theta + \sin^2\theta = 1\), we get: \(x^2 + y^2 = 1\) The identity is successfully proved.
In simple words: Substitute \(x\sin\theta\) into the first equation to simplify it and find that \(x = \cos\theta\) and \(y = \sin\theta\). Squaring and adding these values naturally yields 1.

Exam Tip: Splitting the powers of sine and cosine in the first step is the key to unlocking this algebra problem.

Section D: (4 Marks)

Question 12. If tan ϴ + sin ϴ = m and tan ϴ - sin ϴ = n, then prove that m² – n² = 4 \(\sqrt{mn}\)
Answer: Let us evaluate the Left Hand Side (LHS) of our equation: \(\text{LHS} = m^2 - n^2 = (m + n)(m - n)\) Substituting the given expressions for \(m\) and \(n\): \(m + n = (\tan\theta + \sin\theta) + (\tan\theta - \sin\theta) = 2\tan\theta\) \(m - n = (\tan\theta + \sin\theta) - (\tan\theta - \sin\theta) = 2\sin\theta\) Therefore, we have: \(\text{LHS} = (2\tan\theta)(2\sin\theta) = 4\tan\theta \sin\theta\) — (1) Now, let us evaluate the Right Hand Side (RHS): \(\text{RHS} = 4\sqrt{mn}\) Substituting the values of \(m\) and \(n\): \(mn = (\tan\theta + \sin\theta)(\tan\theta - \sin\theta) = \tan^2\theta - \sin^2\theta\) Converting tangent to sine and cosine: \(mn = \frac{\sin^2\theta}{\cos^2\theta} - \sin^2\theta\) Factoring out \(\sin^2\theta\): \(mn = \sin^2\theta \left(\frac{1}{\cos^2\theta} - 1\right)\) Since \(\frac{1}{\cos^2\theta} = \sec^2\theta\), we have: \(mn = \sin^2\theta (\sec^2\theta - 1)\) Using the identity \(\sec^2\theta - 1 = \tan^2\theta\): \(mn = \sin^2\theta \tan^2\theta\) Taking the square root: \(\sqrt{mn} = \sin\theta \tan\theta\) Thus, the RHS becomes: \(\text{RHS} = 4\sin\theta \tan\theta\) — (2) Comparing equations (1) and (2), we find that LHS = RHS. The relation is verified.
In simple words: Show that both sides of the equation simplify to \(4\tan\theta\sin\theta\) independently to establish the proof.

Exam Tip: Factoring out \(\sin^2\theta\) under the square root is the crucial step to simplify the product \(mn\).

Question 13. If sin ϴ + cos ϴ = p and sec ϴ + cosec ϴ = q , then prove that q (p² – 1) = 2p
Answer: Let us evaluate the Left Hand Side (LHS) of the given equation: \(\text{LHS} = q(p^2 - 1)\) First, we find \(p^2 - 1\): \(p^2 = (\sin\theta + \cos\theta)^2 = \sin^2\theta + \cos^2\theta + 2\sin\theta \cos\theta\) Since \(\sin^2\theta + \cos^2\theta = 1\), this becomes: \(p^2 = 1 + 2\sin\theta \cos\theta\)
\(\implies p^2 - 1 = 2\sin\theta \cos\theta\) — (1) Now, let us express \(q\) in terms of sine and cosine: \(q = \sec\theta + \csc\theta = \frac{1}{\cos\theta} + \frac{1}{\sin\theta}\) \(q = \frac{\sin\theta + \cos\theta}{\sin\theta \cos\theta}\) Substituting \(\sin\theta + \cos\theta = p\): \(q = \frac{p}{\sin\theta \cos\theta}\) — (2) Now, substituting equations (1) and (2) into our LHS expression: \(\text{LHS} = \left(\frac{p}{\sin\theta \cos\theta}\right)(2\sin\theta \cos\theta)\) Cancelling the common term \(\sin\theta \cos\theta\) from the numerator and denominator: \(\text{LHS} = 2p = \text{RHS}\) The equation is proved.
In simple words: Square \(p\) to show \(p^2 - 1\) equals \(2\sin\theta\cos\theta\). Write \(q\) using sine and cosine, and multiply them together to get \(2p\).

Exam Tip: Expressing secant and cosecant in terms of sine and cosine simplifies calculations when dealing with mixed trigonometric expressions.

Question 14. If secϴ = x + \(\frac{1}{4x}\), prove that secϴ + tanϴ = 2x or \(\frac{1}{2x}\)
Answer: We use the standard identity \(\tan^2\theta = \sec^2\theta - 1\). Substituting the given value of \(\sec\theta\): \(\tan^2\theta = \left(x + \frac{1}{4x}\right)^2 - 1\) \(\tan^2\theta = x^2 + \frac{1}{16x^2} + 2(x)\left(\frac{1}{4x}\right) - 1\) \(\tan^2\theta = x^2 + \frac{1}{16x^2} + \frac{1}{2} - 1\) \(\tan^2\theta = x^2 + \frac{1}{16x^2} - \frac{1}{2}\) This can be written as a perfect square: \(\tan^2\theta = \left(x - \frac{1}{4x}\right)^2\) Taking the square root on both sides: \(\tan\theta = \pm \left(x - \frac{1}{4x}\right)\) Now, let us analyze the two cases:
(i) Taking the positive value: \(\tan\theta = x - \frac{1}{4x}\) \(\sec\theta + \tan\theta = \left(x + \frac{1}{4x}\right) + \left(x - \frac{1}{4x}\right) = 2x\)
(ii) Taking the negative value: \(\tan\theta = -\left(x - \frac{1}{4x}\right) = -x + \frac{1}{4x}\) \(\sec\theta + \tan\theta = \left(x + \frac{1}{4x}\right) + \left(-x + \frac{1}{4x}\right) = \frac{2}{4x} = \frac{1}{2x}\) Hence, \(\sec\theta + \tan\theta = 2x\) or \(\frac{1}{2x}\).
In simple words: Square \(\sec\theta\) and subtract 1 to find \(\tan\theta\), which has two square root values. Adding each to \(\sec\theta\) yields the two required outcomes.

Exam Tip: Do not forget the \(\pm\) sign when taking the square root of a perfect square in trigonometric proofs.

Question 15. If A + B = 90⁰, prove that \(\sqrt{\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A}} = \tan A\)
Answer: Since we are given that \(A + B = 90^\circ\), we have \(B = 90^\circ - A\). Using complementary angle relations, we can substitute: \(\tan B = \tan(90^\circ - A) = \cot A\) \(\cot B = \cot(90^\circ - A) = \tan A\) \(\sec B = \sec(90^\circ - A) = \csc A\) \(\sin B = \sin(90^\circ - A) = \cos A\) Now, let us substitute these into the Left Hand Side (LHS) of the equation: \(\text{LHS} = \sqrt{\frac{\tan A \cot A + \tan A \tan A}{\sin A \csc A} - \frac{\cos^2 A}{\cos^2 A}}\) Since \(\tan A \cot A = 1\), \(\sin A \csc A = 1\), and \(\frac{\cos^2 A}{\cos^2 A} = 1\), the expression simplifies to: \(\text{LHS} = \sqrt{\frac{1 + \tan^2 A}{1} - 1}\) \(\text{LHS} = \sqrt{1 + \tan^2 A - 1}\) \(\text{LHS} = \sqrt{\tan^2 A} = \tan A = \text{RHS}\) The identity is proved.
In simple words: Replace \(B\) with \(90^\circ - A\) and simplify the complementary ratios. The expression inside the square root reduces to \(\tan^2 A\), which is equal to \(\tan A\).

Exam Tip: Be sure to write the formulas for the complementary conversions you make in the first step to get full marks.

Question 16. (sec A + tan A)(sec B +tan B)(sec C + tan C)=(sec A – tan A)(sec B –tan B)(sec C – tan C) Prove that each of the side is equal to ± 1.
Answer: Let us assume the value of each side is equal to \(x\): \(x = (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C)\) — (1) \(x = (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)\) — (2) Multiplying equation (1) by equation (2): \(x^2 = \left[(\sec A + \tan A)(\sec A - \tan A)\right] \left[(\sec B + \tan B)(\sec B - \tan B)\right] \left[(\sec C + \tan C)(\sec C - \tan C)\right]\) Using the algebraic identity \((a + b)(a - b) = a^2 - b^2\): \(x^2 = (\sec^2 A - \tan^2 A)(\sec^2 B - \tan^2 B)(\sec^2 C - \tan^2 C)\) Since \(\sec^2\theta - \tan^2\theta = 1\), we substitute this identity: \(x^2 = (1)(1)(1) = 1\) Taking the square root on both sides: \(x = \pm 1\) Therefore, each side is equal to \(\pm 1\).
In simple words: If we multiply both sides of the equation together, we can use the difference of squares identity. Since \(\sec^2\theta - \tan^2\theta = 1\), the product equals 1, meaning each side is \(\pm 1\).

Exam Tip: Multiplying the two equal sides is a very elegant algebraic method to solve this system of identities.

Question 17. In a acute angled triangle ABC, if sin (A + B – C) = \(\frac{1}{2}\) and cos (B + C – A) = \(\frac{1}{\sqrt{2}}\), find \(\angle A\), \(\angle B\) and \(\angle C\).
Answer: Since \(ABC\) is a triangle, the sum of its interior angles is: \(A + B + C = 180^\circ\) — (1) We are given: \(\sin(A + B - C) = \frac{1}{2}\) Since \(\sin 30^\circ = \frac{1}{2}\), we get: \(A + B - C = 30^\circ\) — (2) We are also given: \(\cos(B + C - A) = \frac{1}{\sqrt{2}}\) Since \(\cos 45^\circ = \frac{1}{\sqrt{2}}\), we get: \(B + C - A = 45^\circ\) — (3) Let us add equations (1) and (2): \((A + B + C) + (A + B - C) = 180^\circ + 30^\circ\) \(2A + 2B = 210^\circ \implies A + B = 105^\circ\) — (4) Substituting equation (4) into equation (1): \(105^\circ + C = 180^\circ \implies C = 75^\circ\) Now, substituting the value of \(C = 75^\circ\) into equation (3): \(B + 75^\circ - A = 45^\circ \implies A - B = 30^\circ\) — (5) By solving equations (4) and (5) simultaneously:
• Adding (4) and (5): \(2A = 135^\circ \implies A = 67.5^\circ\)
• Substituting \(A\) into (5): \(67.5^\circ - B = 30^\circ \implies B = 37.5^\circ\) Thus, the angles are \(\angle A = 67.5^\circ\), \(\angle B = 37.5^\circ\), and \(\angle C = 75^\circ\).
In simple words: Set up a system of equations by matching the sine and cosine values to standard angles, then use the angle sum property of triangles to solve for \(A\), \(B\), and \(C\).

Exam Tip: Be sure to include the triangle sum property equation (\(A + B + C = 180^\circ\)) as it is crucial to solving the three variables.

Question 18. If sin A = \(\frac{1}{\sqrt{5}}\) and sin B = \(\frac{1}{\sqrt{10}}\), find the values of cos A and cos B. Hence using the formula cos (A + B) = cos A cos B – sin A sin B, show that (A + B) = 45⁰.
Answer: We first find the values of \(\cos A\) and \(\cos B\) using \(\cos\theta = \sqrt{1 - \sin^2\theta}\): \(\cos A = \sqrt{1 - \left(\frac{1}{\sqrt{5}}\right)^2} = \sqrt{1 - \frac{1}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}\) \(\cos B = \sqrt{1 - \left(\frac{1}{\sqrt{10}}\right)^2} = \sqrt{1 - \frac{1}{10}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}\) Now, substituting these values into the given formula: \(\cos(A + B) = \cos A \cos B - \sin A \sin B\) \(\cos(A + B) = \left(\frac{2}{\sqrt{5}}\right)\left(\frac{3}{\sqrt{10}}\right) - \left(\frac{1}{\sqrt{5}}\right)\left(\frac{1}{\sqrt{10}}\right)\) \(\cos(A + B) = \frac{6}{\sqrt{50}} - \frac{1}{\sqrt{50}} = \frac{5}{\sqrt{50}}\) Since \(\sqrt{50} = 5\sqrt{2}\): \(\cos(A + B) = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}\) Since we know that \(\cos 45^\circ = \frac{1}{\sqrt{2}}\), we can equate the angles: \(A + B = 45^\circ\). The identity is proved.
In simple words: Find the cosines of both angles, plug them into the compound angle identity, and show that the result is \(\frac{1}{\sqrt{2}}\), which matches \(45^\circ\).

Exam Tip: Be sure to keep the denominators in their radical form (\(\sqrt{5}\) and \(\sqrt{10}\)) to make the final simplification easier.

Question 19. Prove that : sec²ϴ - \(\frac{sin²\theta - 2sin⁴\theta}{2cos⁴\theta - cos²\theta}\) = 1
Answer: Let us simplify the fraction in the given expression: \(\text{Fraction} = \frac{\sin^2\theta - 2\sin^4\theta}{2\cos^4\theta - \cos^2\theta}\) Factoring out \(\sin^2\theta\) from the numerator and \(\cos^2\theta\) from the denominator: \(\text{Fraction} = \frac{\sin^2\theta(1 - 2\sin^2\theta)}{\cos^2\theta(2\cos^2\theta - 1)}\) We can rewrite \(\sin^2\theta\) in the numerator as \(1 - \cos^2\theta\): \(1 - 2\sin^2\theta = 1 - 2(1 - \cos^2\theta) = 1 - 2 + 2\cos^2\theta = 2\cos^2\theta - 1\) Substituting this back into the fraction: \(\text{Fraction} = \frac{\sin^2\theta(2\cos^2\theta - 1)}{\cos^2\theta(2\cos^2\theta - 1)}\) Cancelling the common term \((2\cos^2\theta - 1)\): \(\text{Fraction} = \frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta\) Now substitute this back into our original expression: \(\text{LHS} = \sec^2\theta - \tan^2\theta\) Since \(\sec^2\theta - \tan^2\theta = 1\), we have: \(\text{LHS} = 1 = \text{RHS}\) The identity is proved.
In simple words: Factor the fraction to cancel out the complex terms, reducing it to \(\tan^2\theta\). Subtracting this from \(\sec^2\theta\) yields the standard identity value of 1.

Exam Tip: Converting the bracketed terms to a single common function, like cosine, is a safe way to verify if they can be cancelled.

Question 20. If a cos ϴ - b sin ϴ = c, prove that a sin ϴ + b cos ϴ = ± \(\sqrt{a^2 + b^2 - c^2}\).
Answer: Let us denote the required expression as \(x\): \(a\sin\theta + b\cos\theta = x\) — (1) We are given: \(a\cos\theta - b\sin\theta = c\) — (2) Let us square and add equations (1) and (2): \(x^2 + c^2 = (a\sin\theta + b\cos\theta)^2 + (a\cos\theta - b\sin\theta)^2\) Expanding both squared expressions: \(x^2 + c^2 = (a^2\sin^2\theta + b^2\cos^2\theta + 2ab\sin\theta\cos\theta) + (a^2\cos^2\theta + b^2\sin^2\theta - 2ab\sin\theta\cos\theta)\) Cancelling out the term \(2ab\sin\theta\cos\theta\): \(x^2 + c^2 = a^2\sin^2\theta + a^2\cos^2\theta + b^2\cos^2\theta + b^2\sin^2\theta\) Grouping terms: \(x^2 + c^2 = a^2(\sin^2\theta + \cos^2\theta) + b^2(\cos^2\theta + \sin^2\theta)\) Since \(\sin^2\theta + \cos^2\theta = 1\), this simplifies to: \(x^2 + c^2 = a^2(1) + b^2(1)\) \(x^2 + c^2 = a^2 + b^2\) Isolating \(x^2\): \(x^2 = a^2 + b^2 - c^2\) Taking the square root on both sides: \(x = \pm \sqrt{a^2 + b^2 - c^2}\) Substituting back the value of \(x\): \(a\sin\theta + b\cos\theta = \pm \sqrt{a^2 + b^2 - c^2}\). Hence, the identity is proved.
In simple words: Square both expressions and add them together. This cancels the middle terms and simplifies to \(a^2 + b^2\) using fundamental identities. Solve for the missing expression by taking the square root.

Exam Tip: Squaring and adding is a highly effective method whenever you need to find the sum of complementary coefficient terms.