$1 - 2sin^2 \left ( \dfrac{\pi}{4} - \dfrac{\theta}{2} \right ) = sin \theta$ Solution: LHS $= 1 - 2sin^2 \left ( \dfrac{\pi}{4} - \dfrac{\theta}{2} \right )$ [ Using formula, cos2A = 1 - 2sin²A ] $= cos 2\left ( \dfrac{\pi}{4} - \dfrac{\theta}{2} \right )$ $= cos \left ( \dfrac{\pi}{4} × 2 - \dfrac{\theta}{2} × 2 \right )$ $= cos \…

Prove that: $(cosA - cosB)^2 + (sinA - sinB)^2 = 4sin^2 \left ( \dfrac{A-B}{2} \right )$ LHS $= (cosA - cosB)^2 + (sinA - sinB)^2$ [ Expanding the formulae ] $= cos^2A + cos^2B - 2cosAcosB + sin^2A + sin^2B - 2sinAsinB$ $= (cos^2A + sin^2A )+ (cos^2B + sin^2B )- 2(sinAsinB + cosAcosB)$ [Using trigonometric identity: cos²A + sin²A = …

Prove the following trigonometric identity: $tan \left ( \dfrac{\pi}{4} - \dfrac{A}{2} \right ) = \dfrac{ cos \frac{A}{2} - sin\frac{A}{2} } {cos \frac{A}{2} + sin \frac{A}{2}} = \dfrac{cosA}{1+sinA}$ Solution: Left Side $= tan ( \frac{\pi}{4} - \frac{A}{2} )$ $= tan (\frac{180°}{4} - \frac{A}{2} )$ $= tan (45° - \frac{A}{2})$ $= \d…

Solution: Given, $sin \frac{\theta}{3} = \frac{1}{2} \left ( m + \dfrac{1}{m} \right )$ Now, $sin \theta$ $= 3sin \frac{\theta}{3} - 4sin^3 \frac{\theta}{3} $ $= sin \frac{\theta}{3} \left ( 3 - 4 sin^2 \frac{\theta}{3} \right )$ $= sin \frac{\theta}{3} \left ( 3 - 4 \left [ \dfrac{1}{2} \left ( m + \frac{1}{m} \right ) \right ]^2 \…

1. If $cos \frac{ \theta }{2}= \dfrac{4}{5}$, find the value of: Solution: Here, $cos \theta = 2 cos^2 \frac{\theta}{2} -1$ $= 2 \left ( \dfrac{4}{5} \right )^2 - 1$ $= 2 × \dfrac{16}{25} -1$ $= \dfrac{32 - 25}{25}$ $= \dfrac{7}{25}$ Now, a) $sin \theta$ Solution: $= \sqrt{…

Solution: RHS $= cot \frac{A}{2} - tan \frac{A}{2}$ $= \dfrac{cos \frac{A}{2} }{sin \frac{A}{2}} - \dfrac{sin \frac{A}{2} }{cos \frac{A}{2}}$ $= \dfrac{cos^2 \frac{A}{2} - sin^2 \frac{A}{2}}{sin \frac{A}{2} cos\frac{A}{2}$ $= \dfrac{cosA}{sin \frac{A}{2} cos\frac{A}{2} } × \dfrac{2}{2}$ $= \dfrac{2cosA}{2 cos\frac{A}{2} sin\frac{A}{…

Solution: LHS $= \dfrac{sin 2\theta}{1 + cos 2\theta} × \dfrac{cos \theta}{1 + cos\theta} $ $= \dfrac{sin 2\theta}{1 + 2cos^2 \theta - 1} × \dfrac{cos \theta}{1 + cos\theta}$ $= \dfrac{2sin \theta cos\theta}{2cos^2 \theta} × \dfrac{cos \theta}{1 + cos\theta}$ $= \dfrac{sin \theta}{cos \theta} × \dfrac{cos\theta}{1 + cos\theta}$ $= \…

Prove the following trigonometric identity: $\dfrac{2sinA + sin2A}{2sinA - sin2A} = cot^2 \frac{A}{2}$ Solution: LHS $= \dfrac{2sinA + sin2A}{2sinA - sin2A}$ [ Using sin2A = 2sinAcosA ] $= \dfrac{2sinA + 2sinAcosA}{2sinA - 2sinAcosA}$ $= \dfrac{2sinA(1+cosA)}{2sinA(1-cosA)}$ $= \dfrac{1 + cosA}{1-cosA}$ [ Using sub-multiple angle fo…

Question:Prove that: $\dfrac{1 + cos \alpha}{1 - cos \alpha} = cot^2 \frac{ \alpha}{2}$ Solution : Taking LHS $= \dfrac{1 + cos \alpha}{1 - cos \alpha}$ Using sub-multiple formula of cosA and trigonometric identity of 1. $= \dfrac{(sin^2 \frac{\alpha}{2} + cos^2 \frac{\alpha}{2}) + (cos^2 \frac{\alpha}{2} - sin^2 \frac{\alpha}{2})}{…

Question: Prove that: $\dfrac{1 + cos \alpha}{sin \alpha} = cot \frac{\alpha}{2}$ Solution: Taking LHS $= \dfrac{1 + cos \alpha}{sin \alpha}$ $= \dfrac{(sin^2 \frac{\alpha }{2} + cos^2\frac{\alpha}{2}) + (cos^2 \frac{\alpha}{2} - sin^2 \frac{\alpha}{2})}{sin \alpha}$ $= \dfrac{2cos^2 \frac{\alpha}{2}}{2 sin \frac{alpha}{2} cos \frac…

Question: Prove that: $cos A = \dfrac{1 - tan^2 \frac{A}{2}}{1 + tan^2 \frac{A}{2}}$ Solution: To prove: $cos A = \dfrac{1 - tan^2 \frac{A}{2}}{1 + tan^2 \frac{A}{2}}$ Taking LHS, $= cos A$ $= \dfrac{cos A}{1}$ $= \dfrac{cos^2 \frac{A}{2} - sin^2 \frac{A}{2}}{cos^2 \frac{A}{2} + sin^2 \frac{A}{2}}$ [Dividing numerator and denominato…

Questions: Prove the following simple identity: $cos \theta = 4 cos^3 \frac{ \theta}{3} - 3 cos \frac{ \theta}{3}$ Solution: To prove: $cos \theta = 4 cos^3 \frac{ \theta}{3} - 3 cos \frac{ \theta}{3}$ Taking LHS $cos \theta$ $= cos \left ( \frac{ \theta}{3} + \frac{2 \theta}{3} \right )$ $= cos \frac{ \theta}{3} × cos \frac{ 2\thet…

Question: Prove the following simple trigonometric identity: $\left ( sin \frac{A}{2} - cos \frac{A}{2} \right )^2 = 1 - sin A$ Solution: To prove: $\left ( sin \frac{A}{2} - cos \frac{A}{2} \right )^2 = 1 - sin A$ Taking LHS, $\left ( sin \frac{A}{2} - cos \frac{A}{2} \right )^2$ $= (sin \frac{A}{2} )^2 - 2 × sin \frac{A}{2} × cos …