CBSE Class 10 Mathematics Some Applications of Trigonometry Worksheet Set 10

Question 1. If xcosθ – ysinθ = a, xsinθ + ycos θ = b, prove that x² + y² = a² + b².
Answer:
We are given:
\[ x\cos\theta - y\sin\theta = a \]
\[ x\sin\theta + y\cos\theta = b \]
Squaring both equations:
\[ (x\cos\theta - y\sin\theta)^2 = a^2 \]

\[ \implies x^2\cos^2\theta + y^2\sin^2\theta - 2xy\sin\theta\cos\theta = a^2 \quad \text{--- (Equation 1)} \]
\[ (x\sin\theta + y\cos\theta)^2 = b^2 \]

\[ \implies x^2\sin^2\theta + y^2\cos^2\theta + 2xy\sin\theta\cos\theta = b^2 \quad \text{--- (Equation 2)} \]
Adding Equation 1 and Equation 2:
\[ (x^2\cos^2\theta + y^2\sin^2\theta - 2xy\sin\theta\cos\theta) + (x^2\sin^2\theta + y^2\cos^2\theta + 2xy\sin\theta\cos\theta) = a^2 + b^2 \]
The cross-product terms cancel out:
\[ x^2\cos^2\theta + x^2\sin^2\theta + y^2\sin^2\theta + y^2\cos^2\theta = a^2 + b^2 \]
\[ x^2(\cos^2\theta + \sin^2\theta) + y^2(\sin^2\theta + \cos^2\theta) = a^2 + b^2 \]
Using the identity \( \sin^2\theta + \cos^2\theta = 1 \):
\[ x^2(1) + y^2(1) = a^2 + b^2 \]
\[ x^2 + y^2 = a^2 + b^2 \]
Hence proved.
In simple words: Square both equations. When you add them together, the middle terms cancel out, and grouping the remaining terms simplifies to twice the standard identity, which equals 1.

Exam Tip: Squaring and adding is the most standard algebraic technique to eliminate trigonometric variables when they are multiplied by sine and cosine.

Question 2. Prove that sec²θ + cosec²θ can never be less than 2.
Answer:
Using the identities \( \sec^2\theta = 1 + \tan^2\theta \) and \( \csc^2\theta = 1 + \cot^2\theta \):
\[ \sec^2\theta + \csc^2\theta = (1 + \tan^2\theta) + (1 + \cot^2\theta) = 2 + \tan^2\theta + \cot^2\theta \]
Since \( \tan\theta \) and \( \cot\theta \) are reciprocals, \( \tan\theta \cdot \cot\theta = 1 \). We can write:
\[ 2 + \tan^2\theta + \cot^2\theta = (\tan\theta - \cot\theta)^2 + 2\tan\theta\cot\theta + 2 \]
\[ = (\tan\theta - \cot\theta)^2 + 2(1) + 2 = (\tan\theta - \cot\theta)^2 + 4 \]
Since the square of any real number is always non-negative (\( \ge 0 \)):
\[ (\tan\theta - \cot\theta)^2 \ge 0 \]
\[ \implies (\tan\theta - \cot\theta)^2 + 4 \ge 4 \]
Therefore, \( \sec^2\theta + \csc^2\theta \ge 4 \), which means it can never be less than 4 (and hence, never less than 2).
In simple words: Express secant and cosecant in terms of tangent and cotangent. Since any squared real number must be zero or positive, the expression is always at least 4.

Exam Tip: Remember that for any positive numbers \( a \) and \( b \), the arithmetic mean is greater than or equal to the geometric mean (\( a + b \ge 2\sqrt{ab} \)). This helps you find minimum values easily.

Question 3. If sinϕ = 1/2, show that 3cosϕ-4cos³ϕ = 0.
Answer:
We are given:
\[ \sin\phi = \frac{1}{2} \]
Since we know that \( \sin 30^\circ = \frac{1}{2} \):

\[ \implies \phi = 30^\circ \]
Now, we substitute \( \phi = 30^\circ \) into the given expression:
\[ E = 3\cos(30^\circ) - 4\cos^3(30^\circ) \]
We know that \( \cos 30^\circ = \frac{\sqrt{3}}{2} \):
\[ E = 3\left(\frac{\sqrt{3}}{2}\right) - 4\left(\frac{\sqrt{3}}{2}\right)^3 \]
\[ E = \frac{3\sqrt{3}}{2} - 4\left(\frac{3\sqrt{3}}{8}\right) \]
\[ E = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} = 0 \]
Hence proved.
In simple words: Find the angle whose sine is 1/2, which is 30 degrees. Plug this angle into the second expression to verify that it simplifies to zero.

Exam Tip: You can also prove this by using the triple-angle formula \( \cos 3\phi = 4\cos^3\phi - 3\cos\phi \), which means \( 3\cos\phi - 4\cos^3\phi = -\cos 3\phi \). For \( \phi = 30^\circ \), this equals \( -\cos 90^\circ = 0 \).

Question 4. If 7sin²ϕ + 3cos²ϕ = 4, show that tanϕ = 1/√3.
Answer:
We are given the equation:
\[ 7\sin^2\phi + 3\cos^2\phi = 4 \]
Using the identity \( \sin^2\phi + \cos^2\phi = 1 \), we can multiply the constant 4 by this identity:
\[ 7\sin^2\phi + 3\cos^2\phi = 4(\sin^2\phi + \cos^2\phi) \]
\[ 7\sin^2\phi + 3\cos^2\phi = 4\sin^2\phi + 4\cos^2\phi \]
Subtracting terms on both sides:
\[ 7\sin^2\phi - 4\sin^2\phi = 4\cos^2\phi - 3\cos^2\phi \]
\[ 3\sin^2\phi = \cos^2\phi \]
Divide both sides by \( \cos^2\phi \):
\[ \frac{3\sin^2\phi}{\cos^2\phi} = 1 \]
\[ 3\tan^2\phi = 1 \]
\[ \tan^2\phi = \frac{1}{3} \]
Taking the square root (for acute angle \( \phi \)):
\[ \tan\phi = \frac{1}{\sqrt{3}} \]
Hence proved.
In simple words: Rewrite the right side of the equation using the basic sine-cosine identity. Group the sines and cosines together to get a tangent equation, and solve for tangent.

Exam Tip: Dividing the entire equation by \( \cos^2\phi \) at the beginning is another very popular and direct method to solve this.

Question 5. If cosϕ+sinϕ = √2 cosϕ, prove that cosϕ - sinϕ = √2 sin ϕ.
Answer:
We are given:
\[ \cos\phi + \sin\phi = \sqrt{2}\cos\phi \]
Square both sides:
\[ (\cos\phi + \sin\phi)^2 = (\sqrt{2}\cos\phi)^2 \]
\[ \cos^2\phi + \sin^2\phi + 2\cos\phi\sin\phi = 2\cos^2\phi \]
\[ 2\cos\phi\sin\phi = 2\cos^2\phi - \cos^2\phi - \sin^2\phi = \cos^2\phi - \sin^2\phi \]
Using the identity \( \cos^2\phi - \sin^2\phi = (\cos\phi - \sin\phi)(\cos\phi + \sin\phi) \):
\[ 2\cos\phi\sin\phi = (\cos\phi - \sin\phi)(\cos\phi + \sin\phi) \]
Substitute \( \cos\phi + \sin\phi = \sqrt{2}\cos\phi \):
\[ 2\cos\phi\sin\phi = (\cos\phi - \sin\phi)(\sqrt{2}\cos\phi) \]
Assuming \( \cos\phi \neq 0 \), divide both sides by \( \sqrt{2}\cos\phi \):
\[ \cos\phi - \sin\phi = \frac{2\cos\phi\sin\phi}{\sqrt{2}\cos\phi} = \sqrt{2}\sin\phi \]
Hence proved.
In simple words: Square both sides of the equation. Simplify the algebraic terms, factor them using the difference of squares, and substitute the given equation back to find the result.

Exam Tip: Alternatively, you can isolate \( \sin\phi \) on one side: \( \sin\phi = (\sqrt{2}-1)\cos\phi \). Multiplying both sides by \( \sqrt{2}+1 \) rationalizes the coefficient and yields the proof instantly.

Question 6. If tanA+sinA=m and tanA-sinA=n, show that m² - n² = 4√mn
Answer:
Let us simplify the Left-Hand Side (LHS) of the expression:
\[ m^2 - n^2 = (m + n)(m - n) \]
Here, we add and subtract the given equations:
- \( m + n = (\tan A + \sin A) + (\tan A - \sin A) = 2\tan A \)
- \( m - n = (\tan A + \sin A) - (\tan A - \sin A) = 2\sin A \)
Substituting these values:
\[ m^2 - n^2 = (2\tan A)(2\sin A) = 4\tan A\sin A \]

Now, let us simplify the Right-Hand Side (RHS):
\[ 4\sqrt{mn} = 4\sqrt{(\tan A + \sin A)(\tan A - \sin A)} \]
\[ = 4\sqrt{\tan^2 A - \sin^2 A} \]
Express tangent in terms of sine and cosine:
\[ = 4\sqrt{\frac{\sin^2 A}{\cos^2 A} - \sin^2 A} \]
Factor out \( \sin^2 A \) inside the square root:
\[ = 4\sqrt{\sin^2 A \left(\frac{1}{\cos^2 A} - 1\right)} \]
Using the identity \( \sec^2 A - 1 = \tan^2 A \):
\[ = 4\sqrt{\sin^2 A \cdot \tan^2 A} = 4\sin A\tan A \]
Since \( \text{LHS} = \text{RHS} \), we have \( m^2 - n^2 = 4\sqrt{mn} \).
Hence proved.
In simple words: Factor the left side as a difference of squares to find that it equals 4 times tangent times sine. Do the same with the right side product to show both sides are identical.

Exam Tip: Proving that \( \tan^2 A - \sin^2 A = \tan^2 A\sin^2 A \) is the core identity step required to solve this problem.

Question 7. If secA = x + 1/(4x), prove that secA+tanA=2x or 1/(2x).
Answer:
We know the identity:
\[ \tan^2 A = \sec^2 A - 1 \]
Substitute the given value of \( \sec A \):
\[ \tan^2 A = \left(x + \frac{1}{4x}\right)^2 - 1 \]
\[ \tan^2 A = x^2 + \frac{1}{16x^2} + 2(x)\left(\frac{1}{4x}\right) - 1 \]
\[ \tan^2 A = x^2 + \frac{1}{16x^2} + \frac{1}{2} - 1 \]
\[ \tan^2 A = x^2 + \frac{1}{16x^2} - \frac{1}{2} \]
This forms a perfect square expansion:
\[ \tan^2 A = \left(x - \frac{1}{4x}\right)^2 \]
Taking the square root on both sides:
\[ \tan A = \pm \left(x - \frac{1}{4x}\right) \delta \]
Case 1: Taking the positive value of \( \tan A \):
\[ \sec A + \tan A = \left(x + \frac{1}{4x}\right) + \left(x - \frac{1}{4x}\right) = 2x \]
Case 2: Taking the negative value of \( \tan A \):
\[ \sec A + \tan A = \left(x + \frac{1}{4x}\right) - \left(x - \frac{1}{4x}\right) = \frac{2}{4x} = \frac{1}{2x} \]
Hence proved.
In simple words: Find tangent squared by using the standard identity. Squaring the terms reveals a perfect square, which gives two possible values for tangent, leading to the two requested solutions.

Exam Tip: Be sure to write the \( \pm \) sign when taking the square root to evaluate both cases and complete the full proof.

Question 8. If A, B are acute angles and sinA= cosB, then find the value of A+B.
Answer:
Given:
\[ \sin A = \cos B \delta \]
Using the complementary angle identity \( \cos B = \sin(90^\circ - B) \):
\[ \sin A = \sin(90^\circ - B) \]
Comparing the angles on both sides:
\[ A = 90^\circ - B \]
\[ A + B = 90^\circ \]
The value of \( A + B \) is \( 90^\circ \).
In simple words: Since sine of an angle is equal to the cosine of its complement, the two acute angles must add up to 90 degrees.

Exam Tip: For any right-angled triangle, the sum of the two acute angles is always \( 90^\circ \).

Question 9. a)Solve for ϕ, if tan5ϕ = 1.
b)Solve for ϕ if \(\frac{\sin\phi}{1+\cos\phi} + \frac{1+\cos\phi}{\sin\phi} = 4\).
Answer:
a) Solving \( \tan 5\phi = 1 \):
We know that \( \tan 45^\circ = 1 \). Therefore:
\[ 5\phi = 45^\circ \implies \phi = 9^\circ \]

b) Solving the equation:
\[ \frac{\sin\phi}{1+\cos\phi} + \frac{1+\cos\phi}{\sin\phi} = 4 \]
Find a common denominator:
\[ \frac{\sin^2\phi + (1 + \cos\phi)^2}{\sin\phi(1 + \cos\phi)} = 4 \]
\[ \frac{\sin^2\phi + 1 + \cos^2\phi + 2\cos\phi}{\sin\phi(1 + \cos\phi)} = 4 \]
Using \( \sin^2\phi + \cos^2\phi = 1 \):
\[ \frac{1 + 1 + 2\cos\phi}{\sin\phi(1 + \cos\phi)} = 4 \]
\[ \frac{2(1 + \cos\phi)}{\sin\phi(1 + \cos\phi)} = 4 \]
Cancel the common term \( (1 + \cos\phi) \):
\[ \frac{2}{\sin\phi} = 4 \]
\[ \sin\phi = \frac{2}{4} = \frac{1}{2} \]
Since \( \sin 30^\circ = \frac{1}{2} \), we have:
\[ \phi = 30^\circ \]
In simple words: For part (a), equate the angle to 45 degrees. For part (b), combine the fractions using cross-multiplication, simplify using standard identities, and solve for sine to find the angle.

Exam Tip: Be careful with algebra when expanding squares like \( (1 + \cos\phi)^2 \) to ensure you include the middle term \( 2\cos\phi \).

Question 10. If \(\frac{\cos\alpha}{\cos\beta} = m\) and \(\frac{\cos\alpha}{\sin\beta} = n\), show that \((m^2+n^2)\cos^2\beta = n^2\)
Answer:
We are given:
\[ m = \frac{\cos\alpha}{\cos\beta} \implies m^2 = \frac{\cos^2\alpha}{\cos^2\beta} \]
\[ n = \frac{\cos\alpha}{\sin\beta} \implies n^2 = \frac{\cos^2\alpha}{\sin^2\beta} \delta \]
Now, let's simplify the Left-Hand Side (LHS):
\[ \text{LHS} = (m^2 + n^2)\cos^2\beta \]
Substituting the values of \( m^2 \) and \( n^2 \):
\[ = \left( \frac{\cos^2\alpha}{\cos^2\beta} + \frac{\cos^2\alpha}{\sin^2\beta} \right) \cos^2\beta \]
Factor out \( \cos^2\alpha \):
\[ = \cos^2\alpha \left( \frac{1}{\cos^2\beta} + \frac{1}{\sin^2\beta} \right) \cos^2\beta \]
Combine the fractions inside the parentheses:
\[ = \cos^2\alpha \left( \frac{\sin^2\beta + \cos^2\beta}{\cos^2\beta \cdot \sin^2\beta} \right) \cos^2\beta \]
Using the identity \( \sin^2\beta + \cos^2\beta = 1 \):
\[ = \cos^2\alpha \left( \frac{1}{\cos^2\beta \cdot \sin^2\beta} \right) \cos^2\beta \]
Cancel out \( \cos^2\beta \):
\[ = \frac{\cos^2\alpha}{\sin^2\beta} = n^2 = \text{RHS} \]
Hence proved.
In simple words: Substitute the algebraic fractions into the formula. Factoring out cosine squared alpha and combining the fractions simplifies the expression directly to the right side.

Exam Tip: Factoring common terms before adding fractions makes calculations much simpler and prevents algebraic errors.

Question 11. If 7 cosecϕ-3cotϕ = 7, prove that 7cotϕ - 3cosecϕ = 3.
Answer:
Given:
\[ 7\csc\phi - 3\cot\phi = 7 \]
\[ 7\csc\phi - 7 = 3\cot\phi \]
\[ 7(\csc\phi - 1) = 3\cot\phi \]
Multiply both sides by \( (\csc\phi + 1) \):
\[ 7(\csc\phi - 1)(\csc\phi + 1) = 3\cot\phi(\csc\phi + 1) \]
\[ 7(\csc^2\phi - 1) = 3\cot\phi(\csc\phi + 1) \]
Since \( \csc^2\phi - 1 = \cot^2\phi \):
\[ 7\cot^2\phi = 3\cot\phi(\csc\phi + 1) \]
Assuming \( \cot\phi \neq 0 \), divide both sides by \( \cot\phi \):
\[ 7\cot\phi = 3(\csc\phi + 1) \]
\[ 7\cot\phi = 3\csc\phi + 3 \]
\[ 7\cot\phi - 3\csc\phi = 3 \]
Hence proved.
In simple words: Group the constant coefficients on one side, then multiply by the conjugate term to use the standard identity. Factoring and simplifying yields the desired proof.

Exam Tip: Utilizing the identity \( \csc^2\phi - \cot^2\phi = 1 \) is the easiest way to solve equations that have different combinations of cosecant and cotangent.

Question 12. 2(sin⁶ϕ+cos⁶ϕ) – 3(sin⁴ϕ+cos⁴ϕ)+1 = 0
Answer:
Let us simplify the terms inside the parentheses:
1. For \( \sin^6\phi + \cos^6\phi \):
Using the identity \( a^3 + b^3 = (a + b)^3 - 3ab(a + b) \), where \( a = \sin^2\phi \) and \( b = \cos^2\phi \):
\[ \sin^6\phi + \cos^6\phi = (\sin^2\phi + \cos^2\phi)^3 - 3\sin^2\phi\cos^2\phi(\sin^2\phi + \cos^2\phi) \]
\[ = (1)^3 - 3\sin^2\phi\cos^2\phi(1) = 1 - 3\sin^2\phi\cos^2\phi \]

2. For \( \sin^4\phi + \cos^4\phi \):
Using the identity \( a^2 + b^2 = (a + b)^2 - 2ab \), where \( a = \sin^2\phi \) and \( b = \cos^2\phi \):
\[ \sin^4\phi + \cos^4\phi = (\sin^2\phi + \cos^2\phi)^2 - 2\sin^2\phi\cos^2\phi \]
\[ = (1)^2 - 2\sin^2\phi\cos^2\phi = 1 - 2\sin^2\phi\cos^2\phi \delta \]
Now, substitute these into the Left-Hand Side (LHS) of the given expression:
\[ \text{LHS} = 2(1 - 3\sin^2\phi\cos^2\phi) - 3(1 - 2\sin^2\phi\cos^2\phi) + 1 \]
\[ = 2 - 6\sin^2\phi\cos^2\phi - 3 + 6\sin^2\phi\cos^2\phi + 1 \]
The terms with \( \sin^2\phi\cos^2\phi \) cancel out:
\[ = 2 - 3 + 1 = 0 = \text{RHS} \]
Hence proved.
In simple words: Rewrite the higher-power terms using algebraic identities. Substituting the basic identity simplifies the equation, causing all trigonometric variables to cancel out to zero.

Exam Tip: Knowing algebraic identities like sum of cubes and sum of squares is highly useful for simplifying complex trigonometric expressions.

Question 13. (sin⁸A- cos⁸A) = (2sin²A – 1) (1- 2sin²A cos² A)
Answer:
Let us start with the Left-Hand Side (LHS):
\[ \text{LHS} = \sin^8 A - \cos^8 A \]
This is a difference of squares:
\[ = (\sin^4 A)^2 - (\cos^4 A)^2 \]
\[ = (\sin^4 A - \cos^4 A)(\sin^4 A + \cos^4 A) \]
Factor the first term further as a difference of squares:
\[ \sin^4 A - \cos^4 A = (\sin^2 A - \cos^2 A)(\sin^2 A + \cos^2 A) \]
Since \( \sin^2 A + \cos^2 A = 1 \):
\[ = (\sin^2 A - \cos^2 A)(1) = \sin^2 A - \cos^2 A \]
Using the identity \( \cos^2 A = 1 - \sin^2 A \):
\[ = \sin^2 A - (1 - \sin^2 A) = 2\sin^2 A - 1 \]
Now, simplify the second term, \( \sin^4 A + \cos^4 A \):
\[ \sin^4 A + \cos^4 A = (\sin^2 A + \cos^2 A)^2 - 2\sin^2 A\cos^2 A \]
\[ = (1)^2 - 2\sin^2 A\cos^2 A = 1 - 2\sin^2 A\cos^2 A \]
Substitute both simplified parts back into our equation:
\[ \text{LHS} = (2\sin^2 A - 1)(1 - 2\sin^2 A\cos^2 A) = \text{RHS} \]
Hence proved.
In simple words: Apply the difference of squares algebraic formula. Factoring the terms into simpler square terms lets you use the basic identities to complete the proof.

Exam Tip: Always look to break down higher powers (like 8th or 4th powers) into products of squared terms using basic algebraic identities.

Question 14. If tanθ = 5/6 & θ +φ =90° what is the value of cotφ.
Answer:
We are given:
\[ \theta + \phi = 90^\circ \implies \phi = 90^\circ - \theta \]
Now, find the value of \( \cot\phi \):
\[ \cot\phi = \cot(90^\circ - \theta) \]
Using the complementary identity \( \cot(90^\circ - \theta) = \tan\theta \):
\[ \cot\phi = \tan\theta \]
Since we are given \( \tan\theta = \frac{5}{6} \):
\[ \cot\phi = \frac{5}{6} \]
In simple words: Since the two angles are complementary, the cotangent of one is equal to the tangent of the other, which is 5 over 6.

Exam Tip: Complementary relations are very helpful for switching between different trigonometric ratios quickly.

Question 15. What is the value of tanϕ in terms of sinϕ.
Answer:
We know the relation:
\[ \tan\phi = \frac{\sin\phi}{\cos\phi} \]
Using the fundamental Pythagorean identity, we can write \( \cos\phi \) as:
\[ \cos\phi = \sqrt{1 - \sin^2\phi} \]
Substituting this back into the first equation:
\[ \tan\phi = \frac{\sin\phi}{\sqrt{1 - \sin^2\phi}} \]
In simple words: Express tangent as sine over cosine, and then replace cosine with its sine identity using the Pythagorean formula.

Exam Tip: This formula is very useful for changing mixed trigonometric expressions into single-variable equations.

Question 16. If Secϕ+Tanϕ=4 find sin ϕ, cosϕ
Answer:
Given:
\[ \sec\phi + \tan\phi = 4 \quad \text{--- (Equation 1)} \]
We know the identity \( \sec^2\phi - \tan^2\phi = 1 \implies (\sec\phi - \tan\phi)(\sec\phi + \tan\phi) = 1 \). Substituting Equation 1:
\[ \sec\phi - \tan\phi = \frac{1}{4} \quad \text{--- (Equation 2)} \]
Adding Equation 1 and Equation 2:
\[ 2\sec\phi = 4 + \frac{1}{4} = \frac{17}{4} \implies \sec\phi = \frac{17}{8} \]
Since \( \cos\phi = \frac{1}{\sec\phi} \):
\[ \cos\phi = \frac{8}{17} \]
Now, subtract Equation 2 from Equation 1:
\[ 2\tan\phi = 4 - \frac{1}{4} = \frac{15}{4} \implies \tan\phi = \frac{15}{8} \]
Since \( \sin\phi = \tan\phi \cdot \cos\phi \):
\[ \sin\phi = \left(\frac{15}{8}\right) \left(\frac{8}{17}\right) = \frac{15}{17} \]
Thus, \( \sin\phi = \frac{15}{17} \) and \( \cos\phi = \frac{8}{17} \).
In simple words: Use the reciprocal relation of secant and tangent to set up a system of equations. Solving this system gives the individual values for sine and cosine.

Exam Tip: If \( \sec\phi + \tan\phi = x \), then \( \sec\phi - \tan\phi = 1/x \). This reciprocal relation is extremely useful for solving algebraic trig problems.

Question 17. Secϕ+Tanϕ=p, prove that sinϕ = \(\frac{p^2-1}{p^2+1}\)
Answer:
We are given:
\[ \sec\phi + \tan\phi = p \quad \text{--- (Equation 1)} \]
Using the identity \( \sec^2\phi - \tan^2\phi = 1 \implies (\sec\phi - \tan\phi)(\sec\phi + \tan\phi) = 1 \). Substituting Equation 1:
\[ \sec\phi - \tan\phi = \frac{1}{p} \quad \text{--- (Equation 2)} \]
Adding Equation 1 and Equation 2:
\[ 2\sec\phi = p + \frac{1}{p} = \frac{p^2 + 1}{p} \implies \sec\phi = \frac{p^2 + 1}{2p} \implies \cos\phi = \frac{2p}{p^2 + 1} \]
Subtracting Equation 2 from Equation 1:
\[ 2\tan\phi = p - \frac{1}{p} = \frac{p^2 - 1}{p} \implies \tan\phi = \frac{p^2 - 1}{2p} \]
Now, we calculate \( \sin\phi \):
\[ \sin\phi = \tan\phi \cdot \cos\phi = \left(\frac{p^2 - 1}{2p}\right) \left(\frac{2p}{p^2 + 1}\right) = \frac{p^2 - 1}{p^2 + 1} \]
Hence proved.
In simple words: Set up two equations using the secant-tangent identity. Solve for secant and tangent separately, and multiply the tangent by the reciprocal of secant (cosine) to find the sine formula.

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 18. Prove geometrically the value of Sin 60°
Answer:
Consider an equilateral triangle \( ABC \) with side length \( a \). Each angle in this triangle is \( 60^\circ \).
Draw a perpendicular altitude \( AD \) from vertex \( A \) to side \( BC \).
Since \( AD \) is the altitude, it bisects the base \( BC \), so \( BD = \frac{a}{2} \), and \(\angle BAD = 30^\circ\).
In right-angled triangle \( ABD \), using Pythagoras theorem:
\[ AB^2 = AD^2 + BD^2 \]
\[ a^2 = AD^2 + \left(\frac{a}{2}\right)^2 \]
\[ a^2 = AD^2 + \frac{a^2}{4} \]
\[ AD^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4} \]
\[ AD = \frac{\sqrt{3}}{2}a \]
Now, in right-angled triangle \( ABD \), for angle \(\angle B = 60^\circ\):
\[ \sin 60^\circ = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AD}{AB} = \frac{\frac{\sqrt{3}}{2}a}{a} = \frac{\sqrt{3}}{2} \]
Thus, the geometric proof is complete.
In simple words: Draw an equilateral triangle and cut it in half with an altitude line. Applying Pythagoras theorem to the resulting right triangle allows you to find the ratio of the height to the hypotenuse, which is sine of 60 degrees.

Exam Tip: A neat, labeled geometric diagram is essential to secure full marks in geometric derivations.

Question 19. If \(\frac{1 - \tan\theta}{1 + \tan\theta} = \frac{\sqrt{3}-1}{\sqrt{3}+1}\),show that \(\frac{\sin\theta}{\cos 2\theta} = 1\)
Answer:
We are given:
\[ \frac{1 - \tan\theta}{1 + \tan\theta} = \frac{\sqrt{3}-1}{\sqrt{3}+1} \]
Divide the numerator and denominator on the right side by \( \sqrt{3} \):
\[ \frac{1 - \tan\theta}{1 + \tan\theta} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} \]
Comparing both sides, we find:
\[ \tan\theta = \frac{1}{\sqrt{3}} \]
Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \):

\[ \implies \theta = 30^\circ \]
Now, evaluate the given expression with \( \theta = 30^\circ \):
\[ E = \frac{\sin(30^\circ)}{\cos(2 \cdot 30^\circ)} = \frac{\sin 30^\circ}{\cos 60^\circ} \]
Substitute standard values \( \sin 30^\circ = \frac{1}{2} \) and \( \cos 60^\circ = \frac{1}{2} \):
\[ E = \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \]
Hence proved.
In simple words: Compare both sides to find that tangent of the angle is 1 over root 3, which means the angle is 30 degrees. Plug this angle into the fraction to verify that it equals 1.

Exam Tip: Simplifying the right-hand side using fractional division is much faster than performing a long algebraic cross-multiplication.

Question 20. If 2x=secθ and \(\frac{2}{x} = \tan\theta\) ,then find the value of \(2\left(x^2 - \frac{1}{x^2}\right)\).
Answer:
Given equations:
1. \( 2x = \sec\theta \implies x = \frac{\sec\theta}{2} \)
2. \( \frac{2}{x} = \tan\theta \implies \frac{1}{x} = \frac{\tan\theta}{2} \)
Now, substitute these into the expression \( 2\left(x^2 - \frac{1}{x^2}\right) \):
\[ 2\left(x^2 - \frac{1}{x^2}\right) = 2 \left[ \left(\frac{\sec\theta}{2}\right)^2 - \left(\frac{\tan\theta}{2}\right)^2 \right] \]
\[ = 2 \left[ \frac{\sec^2\theta}{4} - \frac{\tan^2\theta}{4} \right] \]
\[ = \frac{2}{4} (\sec^2\theta - \tan^2\theta) \]
Using the standard identity \( \sec^2\theta - \tan^2\theta = 1 \):
\[ = \frac{1}{2} (1) = \frac{1}{2} \]
The value of the expression is \( \frac{1}{2} \).
In simple words: Write \( x \) and \( 1/x \) in terms of secant and tangent. Substituting these into the expression yields a standard identity which simplifies the final answer to one-half.

Exam Tip: Be careful to distinguish whether the question asks for \( 4\left(x^2 - \frac{1}{x^2}\right) \) (which would equal 1) or \( 2\left(x^2 - \frac{1}{x^2}\right) \) (which equals 1/2) to write the correct final value.