Solutions: Introduction to Trigonometry - Test 3

Q1. (b) 1

\((1 + \tan^2 A)(1 - \sin A)(1 + \sin A) = (\sec^2 A)(1 - \sin^2 A)\)
\(= \sec^2 A \cdot \cos^2 A = \frac{1}{\cos^2 A} \cdot \cos^2 A = 1\).

Q2. (b) 9

\(9 \sec^2 A - 9 \tan^2 A = 9(\sec^2 A - \tan^2 A) = 9(1) = 9\).

Q3. (d) \(\cos A\)

\((\sec A + \tan A)(1 - \sin A) = (\frac{1}{\cos A} + \frac{\sin A}{\cos A})(1 - \sin A)\)
\(= \frac{1 + \sin A}{\cos A} \cdot (1 - \sin A) = \frac{1 - \sin^2 A}{\cos A} = \frac{\cos^2 A}{\cos A} = \cos A\).

Q4. (a) Both A and R are true and R is the correct explanation of A.

The identity \(\sin^2 \theta + \cos^2 \theta = 1\) holds for all \(\theta\). Hence, it holds for \(67^\circ\).

Q5. Prove \(\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = 1 - 2\sin^2 \theta\).

LHS \(= \frac{1 - \frac{\sin^2 \theta}{\cos^2 \theta}}{\sec^2 \theta} = \cos^2 \theta (1 - \frac{\sin^2 \theta}{\cos^2 \theta}) = \cos^2 \theta - \sin^2 \theta\)
\(= (1 - \sin^2 \theta) - \sin^2 \theta = 1 - 2\sin^2 \theta = \text{RHS}\).

Q6. Prove \(\sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}} = \csc \theta + \cot \theta\).

LHS \(= \sqrt{\frac{(1 + \cos \theta)(1 + \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)}} = \sqrt{\frac{(1 + \cos \theta)^2}{1 - \cos^2 \theta}}\)
\(= \sqrt{\frac{(1 + \cos \theta)^2}{\sin^2 \theta}} = \frac{1 + \cos \theta}{\sin \theta} = \frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta} = \csc \theta + \cot \theta\).

Q7. Show \(\cos \theta - \sin \theta = \sqrt{2} \sin \theta\).

Given \(\cos \theta + \sin \theta = \sqrt{2} \cos \theta\). Squaring both sides:
\(\cos^2 \theta + \sin^2 \theta + 2\sin \theta \cos \theta = 2 \cos^2 \theta\)
\(1 + 2\sin \theta \cos \theta = 2 \cos^2 \theta \Rightarrow 2\sin \theta \cos \theta = 2 \cos^2 \theta - 1\).
Now consider \((\cos \theta - \sin \theta)^2 = \cos^2 \theta + \sin^2 \theta - 2\sin \theta \cos \theta\)
\(= 1 - (2 \cos^2 \theta - 1) = 2 - 2 \cos^2 \theta = 2(1 - \cos^2 \theta) = 2 \sin^2 \theta\).
Taking square root: \(\cos \theta - \sin \theta = \sqrt{2} \sin \theta\).

Q8. Prove \(\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = \frac{1 + \sin A}{\cos A}\).

LHS \(= \frac{(\tan A + \sec A) - (\sec^2 A - \tan^2 A)}{\tan A - \sec A + 1}\)
\(= \frac{(\tan A + \sec A) - (\sec A - \tan A)(\sec A + \tan A)}{\tan A - \sec A + 1}\)
\(= \frac{(\sec A + \tan A)[1 - (\sec A - \tan A)]}{\tan A - \sec A + 1}\)
\(= \frac{(\sec A + \tan A)(1 - \sec A + \tan A)}{1 - \sec A + \tan A} = \sec A + \tan A = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \frac{1 + \sin A}{\cos A}\).

Q9. Prove \(\frac{\sin \theta - 2\sin^3 \theta}{2\cos^3 \theta - \cos \theta} = \tan \theta\).

LHS \(= \frac{\sin \theta(1 - 2\sin^2 \theta)}{\cos \theta(2\cos^2 \theta - 1)}\)
\(= \tan \theta \cdot \frac{1 - 2(1 - \cos^2 \theta)}{2\cos^2 \theta - 1}\)
\(= \tan \theta \cdot \frac{1 - 2 + 2\cos^2 \theta}{2\cos^2 \theta - 1} = \tan \theta \cdot \frac{2\cos^2 \theta - 1}{2\cos^2 \theta - 1} = \tan \theta\).

Q10. Prove \((\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A\).

LHS \(= (\sin^2 A + \csc^2 A + 2\sin A \csc A) + (\cos^2 A + \sec^2 A + 2\cos A \sec A)\)
\(= (\sin^2 A + \cos^2 A) + \csc^2 A + \sec^2 A + 2(1) + 2(1)\)
\(= 1 + (1 + \cot^2 A) + (1 + \tan^2 A) + 4\)
\(= 1 + 1 + \cot^2 A + 1 + \tan^2 A + 4\)
\(= 7 + \tan^2 A + \cot^2 A = \text{RHS}\).

(i) \(\sin^2 25^\circ + \sin^2 65^\circ = \sin^2 25^\circ + \cos^2(90^\circ - 65^\circ) = \sin^2 25^\circ + \cos^2 25^\circ = 1\).

(ii) \(\sin^2 A + \cos^2 A = 1 \Rightarrow \cos A = \sqrt{1 - \sin^2 A}\).

(iii) LHS \(= \sec^2 A (\sec^2 A - 1) = (1 + \tan^2 A)(\tan^2 A) = \tan^2 A + \tan^4 A = \text{RHS}\).