If 5sinα = 3sin(α + 2β), Find tan(α + β) If \(5\sin\alpha = 3\sin(\alpha + 2\beta)\neq 0\), Find \( \tan(\alpha+\beta) \) Question If \[ 5\sin\alpha=3\sin(\alpha+2\beta)\neq0, \] then \[ \tan(\alpha+\beta) \] is equal to (a) \(2\tan\beta\) (b) \(3\tan\beta\) (c) \(4\tan\beta\) (d) \(6\tan\beta\) Solution Given, \[ 5\sin\alpha=3\sin(\alpha+2\beta) \] Using \[ \sin(\alpha+2\beta) = \sin\alpha\cos2\beta + \cos\alpha\sin2\beta \] Substituting, […]

Value of sin²(π/18) + sin²(π/9) + sin²(7π/18) + sin²(4π/9) Value of \( \sin^2\frac{\pi}{18}+\sin^2\frac{\pi}{9}+\sin^2\frac{7\pi}{18}+\sin^2\frac{4\pi}{9} \) Question Find the value of \[ \sin^2\frac{\pi}{18} +\sin^2\frac{\pi}{9} +\sin^2\frac{7\pi}{18} +\sin^2\frac{4\pi}{9} \] (a) \(1\) (b) \(2\) (c) \(4\) (d) none of these Solution Notice that \[ \frac{7\pi}{18} = \frac{\pi}{2}-\frac{\pi}{9} \] and \[ \frac{4\pi}{9} = \frac{\pi}{2}-\frac{\pi}{18} \] Using \[ \sin\left(\frac{\pi}{2}-\theta\right)=\cos\theta \] we get

The Value of tan x sin(π/2 + x) cos(π/2 – x) The Value of \( \tan x \sin\left(\frac{\pi}{2}+x\right)\cos\left(\frac{\pi}{2}-x\right) \) Question Find the value of \[ \tan x \cdot \sin\left(\frac{\pi}{2}+x\right)\cdot \cos\left(\frac{\pi}{2}-x\right) \] (a) \(1\) (b) \(-1\) (c) \(\frac{1}{2}\sin 2x\) (d) none of these Solution Use the complementary angle identities: \[ \sin\left(\frac{\pi}{2}+x\right)=\cos x \] \[ \cos\left(\frac{\pi}{2}-x\right)=\sin x

Value of (cot(x/2) – tan(x/2))² (1 – 2tanx cot2x) Value of \((\cot\frac{x}{2}-\tan\frac{x}{2})^2(1-2\tan x\cot 2x)\) Question Find the value of \[ \left(\cot\frac{x}{2}-\tan\frac{x}{2}\right)^2 \left(1-2\tan x\cot 2x\right) \] (a) 1 (b) 2 (c) 3 (d) 4 Solution Using the identity \[ \cot\theta-\tan\theta = \frac{\cos^2\theta-\sin^2\theta} {\sin\theta\cos\theta} = \frac{\cos2\theta}{\frac12\sin2\theta} = 2\cot2\theta \] Putting \(\theta=\frac{x}{2}\), \[ \cot\frac{x}{2}-\tan\frac{x}{2} = 2\cot x \]

If sinα + sinβ = a and cosα − cosβ = b, Find tan((α − β)/2) If \( \sin\alpha+\sin\beta=a \) and \( \cos\alpha-\cos\beta=b \), Find \( \tan\frac{\alpha-\beta}{2} \) Question If \[ \sin\alpha+\sin\beta=a \] and \[ \cos\alpha-\cos\beta=b, \] then find \[ \tan\frac{\alpha-\beta}{2} \] (a) \(-\frac{a}{b}\) (b) \(-\frac{b}{a}\) (c) \(\sqrt{a^2+b^2}\) (d) none of these Solution Using the

If tanα = (1 – cosβ)/sinβ, Then Which Relation is True? If \( \tan\alpha=\frac{1-\cos\beta}{\sin\beta} \), Then Which Relation is True? Question If \[ \tan\alpha=\frac{1-\cos\beta}{\sin\beta}, \] then (a) \(\tan 3\alpha=\tan 2\beta\) (b) \(\tan 2\alpha=\tan\beta\) (c) \(\tan 2\beta=\tan\alpha\) (d) none of these Solution Use the standard half-angle identity: \[ \tan\frac{\beta}{2} = \frac{1-\cos\beta}{\sin\beta} \] Given, \[ \tan\alpha =

If 2tanα = 3tanβ, Find tan(α − β) If \(2\tan\alpha = 3\tan\beta\), Find \( \tan(\alpha-\beta) \) Question If \[ 2\tan\alpha = 3\tan\beta \] then \[ \tan(\alpha-\beta) \] is equal to (a) \(\dfrac{\sin2\beta}{5-\cos2\beta}\) (b) \(\dfrac{\cos2\beta}{5-\cos2\beta}\) (c) \(\dfrac{\sin2\beta}{5+\cos2\beta}\) (d) none of these Solution Given, \[ 2\tan\alpha=3\tan\beta \] \[ \tan\alpha=\frac{3}{2}\tan\beta \] Using the identity \[ \tan(\alpha-\beta) = \frac{\tan\alpha-\tan\beta}

If cos x = 1/2(a + 1/a), Find λ Given cos 3x = λ(a³ + 1/a³) If \( \cos x=\frac12\left(a+\frac1a\right) \), Find \( \lambda \) Question If \[ \cos x=\frac12\left(a+\frac1a\right) \] and \[ \cos 3x=\lambda\left(a^3+\frac1{a^3}\right), \] then \( \lambda \) is (a) \( \frac14 \) (b) \( \frac12 \) (c) \( 1 \) (d) none

If in a Triangle ABC, tan A + tan B + tan C = 0, Find cot A cot B cot C If in a ΔABC, tan A + tan B + tan C = 0, Find cot A cot B cot C Question If in a triangle \(ABC\), \[ \tan A+\tan B+\tan C=0 \]

The Value of 2tan(π/10)+3sec(π/10)-4cos(π/10) The Value of \(2\tan\frac{\pi}{10}+3\sec\frac{\pi}{10}-4\cos\frac{\pi}{10}\) Question Find the value of \[ 2\tan\frac{\pi}{10} + 3\sec\frac{\pi}{10} – 4\cos\frac{\pi}{10} \] (a) \(0\) (b) \(\sqrt5\) (c) \(1\) (d) none of these Solution Let \[ \theta=\frac{\pi}{10} \] Then \[ 2\tan\theta+3\sec\theta-4\cos\theta = \frac{2\sin\theta}{\cos\theta} +\frac{3}{\cos\theta} -4\cos\theta \] Taking LCM, \[ = \frac{2\sin\theta+3-4\cos^2\theta}{\cos\theta} \] Using \[ 4\cos^2\theta=2(1+\cos2\theta) \] \[ =