If tan θ = 1/2 and tan ϕ = 1/3, then the value of θ + ϕ is (a) π/6  (b) π  (c) 0  (d) π/4

If tan θ = 1/2 and tan ϕ = 1/3, Find the Value of θ + ϕ If tan θ = 1/2 and tan ϕ = 1/3, Find the Value of θ + ϕ Question: If \[ \tan\theta=\frac{1}{2} \] and \[ \tan\phi=\frac{1}{3} \] then the value of \[ \theta+\phi \] is (a) \(\frac{\pi}{6}\) (b) \(\pi\) […]

If tan θ = 1/2 and tan ϕ = 1/3, then the value of θ + ϕ is (a) π/6  (b) π  (c) 0  (d) π/4 Read More »

If sin (π cos x) = cos (π sin x), then sin 2x = (a) ±3/4  (b) ±4/3  (c) ±1/3  (d) none of these

If sin(π cos x) = cos(π sin x), Find sin 2x If sin(π cos x) = cos(π sin x), Find sin 2x Question: If \[ \sin(\pi\cos x)=\cos(\pi\sin x) \] then \[ \sin2x= \] (a) \(\pm\frac{3}{4}\) (b) \(\pm\frac{4}{3}\) (c) \(\pm\frac{1}{3}\) (d) none of these Solution Using the identity: \[ \cos\theta=\sin\left(\frac{\pi}{2}-\theta\right) \] Therefore, \[ \sin(\pi\cos x) =

If sin (π cos x) = cos (π sin x), then sin 2x = (a) ±3/4  (b) ±4/3  (c) ±1/3  (d) none of these Read More »

If tan θ₁ tan θ₂ = k, then (cos(θ₁ – θ₂) / cos(θ₁ + θ₂)) = (a) (1 + k)/(1 – k)  (b) (1 – k)/(1 + k)  (c) (k + 1)/(k – 1)  (d) (k – 1)/(k + 1)

If tan θ₁ tan θ₂ = k, Find cos(θ₁ − θ₂) / cos(θ₁ + θ₂) If tan θ₁ tan θ₂ = k, Find cos(θ₁ − θ₂) / cos(θ₁ + θ₂) Question: If \[ \tan\theta_1\tan\theta_2=k \] then \[ \frac{\cos(\theta_1-\theta_2)} {\cos(\theta_1+\theta_2)} \] is equal to (a) \(\frac{1+k}{1-k}\) (b) \(\frac{1-k}{1+k}\) (c) \(\frac{k+1}{k-1}\) (d) \(\frac{k-1}{k+1}\) Solution Using the identities:

If tan θ₁ tan θ₂ = k, then (cos(θ₁ – θ₂) / cos(θ₁ + θ₂)) = (a) (1 + k)/(1 – k)  (b) (1 – k)/(1 + k)  (c) (k + 1)/(k – 1)  (d) (k – 1)/(k + 1) Read More »

The value of cos²(π/6 + x) – sin²(π/6 – x) is (a) 1/2 cos 2x  (b) 0  (c) –1/2 cos 2x  (d) 1/2

Find the Value of cos²(π/6 + x) − sin²(π/6 − x) Find the Value of cos²(π/6 + x) − sin²(π/6 − x) Question: The value of \[ \cos^2\left(\frac{\pi}{6}+x\right) – \sin^2\left(\frac{\pi}{6}-x\right) \] is (a) \(\frac{1}{2}\cos2x\) (b) \(0\) (c) \(-\frac{1}{2}\cos2x\) (d) \(\frac{1}{2}\) Solution Use the identities: \[ \cos^2\theta=\frac{1+\cos2\theta}{2} \] and \[ \sin^2\theta=\frac{1-\cos2\theta}{2} \] Therefore, \[ \cos^2\left(\frac{\pi}{6}+x\right) =

The value of cos²(π/6 + x) – sin²(π/6 – x) is (a) 1/2 cos 2x  (b) 0  (c) –1/2 cos 2x  (d) 1/2 Read More »

(cos 10° – sin 10°)/(cos 10° + sin 10°) is equal to (a) tan 55°  (b) cot 55°  (c) – tan 35°  (d) – cot 35°

Find the Value of (cos 10° − sin 10°)/(cos 10° + sin 10°) Find the Value of (cos 10° − sin 10°)/(cos 10° + sin 10°) Question: \[ \frac{\cos10^\circ-\sin10^\circ} {\cos10^\circ+\sin10^\circ} \] is equal to (a) \(\tan55^\circ\) (b) \(\cot55^\circ\) (c) \(-\tan35^\circ\) (d) \(-\cot35^\circ\) Solution Divide numerator and denominator by \[ \cos10^\circ \] Then, \[ \frac{\cos10^\circ-\sin10^\circ} {\cos10^\circ+\sin10^\circ}

(cos 10° – sin 10°)/(cos 10° + sin 10°) is equal to (a) tan 55°  (b) cot 55°  (c) – tan 35°  (d) – cot 35° Read More »

If cot (α + β) = 0, then sin (α + 2β) is equal to (a) sin α  (b) cos 2β  (c) cos α  (d) sin 2α

If cot(α + β) = 0, Find sin(α + 2β) If cot(α + β) = 0, Find sin(α + 2β) Question: If \[ \cot(\alpha+\beta)=0 \] then \[ \sin(\alpha+2\beta) \] is equal to (a) \(\sin\alpha\) (b) \(\cos2\beta\) (c) \(\cos\alpha\) (d) \(\sin2\alpha\) Solution Given, \[ \cot(\alpha+\beta)=0 \] We know that \[ \cot\theta=0 \] when \[ \theta=\frac{\pi}{2} \]

If cot (α + β) = 0, then sin (α + 2β) is equal to (a) sin α  (b) cos 2β  (c) cos α  (d) sin 2α Read More »

If cos P = 1/7 and cos Q = 13/14, where P and Q both are acute angles. Then, the value of P – Q is (a) π/6  (b) π/3  (c) π/4  (d) 5π/12

If cos P = 1/7 and cos Q = 13/14, Find the Value of P − Q If cos P = 1/7 and cos Q = 13/14, Find the Value of P − Q Question: If \[ \cos P=\frac{1}{7} \] and \[ \cos Q=\frac{13}{14} \] where \(P\) and \(Q\) both are acute angles, then the

If cos P = 1/7 and cos Q = 13/14, where P and Q both are acute angles. Then, the value of P – Q is (a) π/6  (b) π/3  (c) π/4  (d) 5π/12 Read More »

If A + B + C = π, then (tan A + tan B + tan C)/(tan A tan B tan C) is equal to (a) tan A tan B tan C  (b) 0  (c) 1  (d) none of these

If A + B + C = π, Find (tan A + tan B + tan C)/(tan A tan B tan C) If A + B + C = π, Find (tan A + tan B + tan C)/(tan A tan B tan C) Question: If \[ A+B+C=\pi \] then \[ \frac{\tan A+\tan B+\tan C}

If A + B + C = π, then (tan A + tan B + tan C)/(tan A tan B tan C) is equal to (a) tan A tan B tan C  (b) 0  (c) 1  (d) none of these Read More »

tan 3A – tan 2A – tan A is equal to (a) tan 3A tan 2A tan A (b) – tan 3A tan 2A tan A (c) tan A tan 2A – tan 2A tan 3A – tan 3A tan A (d) none of these

tan 3A − tan 2A − tan A is Equal To tan 3A − tan 2A − tan A is Equal To Question: \[ \tan3A-\tan2A-\tan A \] is equal to (a) \(\tan3A\tan2A\tan A\) (b) \(-\tan3A\tan2A\tan A\) (c) \(\tan A\tan2A-\tan2A\tan3A-\tan3A\tan A\) (d) none of these Solution Using the identity: \[ \tan(X+Y) = \frac{\tan X+\tan Y} {1-\tan

tan 3A – tan 2A – tan A is equal to (a) tan 3A tan 2A tan A (b) – tan 3A tan 2A tan A (c) tan A tan 2A – tan 2A tan 3A – tan 3A tan A (d) none of these Read More »

If in a ∆ ABC, tan A + tan B + tan C = 6, then cot A cot B cot C = (a) 6  (b) 1  (c) 1/6  (d) none of these

If tan A + tan B + tan C = 6, Find cot A cot B cot C If tan A + tan B + tan C = 6, Find cot A cot B cot C Question: If in a triangle \( \triangle ABC \), \[ \tan A+\tan B+\tan C=6 \] then \[ \cot A\cot

If in a ∆ ABC, tan A + tan B + tan C = 6, then cot A cot B cot C = (a) 6  (b) 1  (c) 1/6  (d) none of these Read More »