Ravi Kant Kumar

If tan(A+B) = x and tan(A−B) = y, Find tan2A and tan2B Question If \[ \tan(A+B)=x \] and \[ \tan(A-B)=y \] find the values of: \[ \tan2A \quad \text{and} \quad \tan2B \] Solution Since, \[ 2A=(A+B)+(A-B) \] \[ \tan2A = \tan[(A+B)+(A-B)] \] \[ = \frac{\tan(A+B)+\tan(A-B)} {1-\tan(A+B)\tan(A-B)} \] \[ = \frac{x+y}{1-xy} \] Therefore, \[ \boxed{\tan2A=\frac{x+y}{1-xy}} \]

If tan A = x tan B, Prove that sin(A−B)/sin(A+B) = (x−1)/(x+1) Question If \[ \tan A=x\tan B \] prove that: \[ \frac{\sin(A-B)}{\sin(A+B)} = \frac{x-1}{x+1} \] Proof Given, \[ \tan A=x\tan B \] \[ \frac{\sin A}{\cos A} = x\frac{\sin B}{\cos B} \] \[ \sin A\cos B = x\cos A\sin B \] Now, \[ \sin(A-B) =

If sin(x+y)/sin(x−y) = (a+b)/(a−b), Show that tanx/tany = a/b Question If \[ \frac{\sin(x+y)}{\sin(x-y)} = \frac{a+b}{a-b} \] show that: \[ \frac{\tan x}{\tan y} = \frac{a}{b} \] Proof \[ \frac{\sin(x+y)}{\sin(x-y)} = \frac{a+b}{a-b} \] Using \[ \sin(x+y)=\sin x\cos y+\cos x\sin y \] and \[ \sin(x-y)=\sin x\cos y-\cos x\sin y \] \[ \frac{\sin x\cos y+\cos x\sin y} {\sin x\cos

Prove that: (tan²2x − tan²x)/(1 − tan²2x tan²x) = tan 3x tan x Question Prove that: \[ \frac{\tan^22x-\tan^2x} {1-\tan^22x\tan^2x} = \tan3x\tan x \] Proof L.H.S. \[ = \frac{(\tan2x-\tan x)(\tan2x+\tan x)} {(1-\tan2x\tan x)(1+\tan2x\tan x)} \] \[ = \frac{\tan2x-\tan x} {1+\tan2x\tan x} \times \frac{\tan2x+\tan x} {1-\tan2x\tan x} \] Using \[ \tan(A-B) = \frac{\tan A-\tan B} {1+\tan A\tan

Prove that: tan 13x − tan 9x − tan 4x = tan 13x tan 9x tan 4x Question Prove that: \[ \tan13x-\tan9x-\tan4x = \tan13x\tan9x\tan4x \] Proof \[ \tan13x = \tan(9x+4x) \] \[ = \frac{\tan9x+\tan4x} {1-\tan9x\tan4x} \] \[ \tan13x(1-\tan9x\tan4x) = \tan9x+\tan4x \] \[ \tan13x-\tan13x\tan9x\tan4x = \tan9x+\tan4x \] \[ \tan13x-\tan9x-\tan4x = \tan13x\tan9x\tan4x \] Hence proved. Next Question

Prove that: tan 36° + tan 9° + tan 36° tan 9° = 1 Question Prove that: \[ \tan36^\circ+\tan9^\circ+\tan36^\circ\tan9^\circ=1 \] Proof \[ \tan(36^\circ+9^\circ) = \frac{\tan36^\circ+\tan9^\circ} {1-\tan36^\circ\tan9^\circ} \] \[ \tan45^\circ = \frac{\tan36^\circ+\tan9^\circ} {1-\tan36^\circ\tan9^\circ} \] \[ 1 = \frac{\tan36^\circ+\tan9^\circ} {1-\tan36^\circ\tan9^\circ} \] \[ 1-\tan36^\circ\tan9^\circ = \tan36^\circ+\tan9^\circ \] \[ \tan36^\circ+\tan9^\circ+\tan36^\circ\tan9^\circ =1 \] Hence proved. Next Question / Full Exercise

Prove that: tan(π/12) + tan(π/6) + tan(π/12)tan(π/6) = 1 Question Prove that: \[ \tan\frac{\pi}{12} + \tan\frac{\pi}{6} + \tan\frac{\pi}{12}\tan\frac{\pi}{6} =1 \] Proof \[ \tan\left(\frac{\pi}{12}+\frac{\pi}{6}\right) = \frac{ \tan\frac{\pi}{12}+\tan\frac{\pi}{6} }{ 1-\tan\frac{\pi}{12}\tan\frac{\pi}{6} } \] \[ \tan\frac{\pi}{4} = \frac{ \tan\frac{\pi}{12}+\tan\frac{\pi}{6} }{ 1-\tan\frac{\pi}{12}\tan\frac{\pi}{6} } \] \[ 1 = \frac{ \tan\frac{\pi}{12}+\tan\frac{\pi}{6} }{ 1-\tan\frac{\pi}{12}\tan\frac{\pi}{6} } \] \[ 1-\tan\frac{\pi}{12}\tan\frac{\pi}{6} = \tan\frac{\pi}{12}+\tan\frac{\pi}{6} \] \[ \tan\frac{\pi}{12}

Prove that: tan 8x − tan 6x − tan 2x = tan 8x tan 6x tan 2x Question Prove that: \[ \tan 8x-\tan 6x-\tan 2x = \tan 8x\tan 6x\tan 2x \] Proof \[ \tan 8x = \tan(6x+2x) \] \[ = \frac{\tan 6x+\tan 2x} {1-\tan 6x\tan 2x} \] \[ \tan 8x(1-\tan 6x\tan 2x) = \tan 6x+\tan

Prove that: tan(A+B)/cot(A−B) = (tan²A − tan²B)/(1 − tan²A tan²B) Question Prove that: \[ \frac{\tan(A+B)}{\cot(A-B)} = \frac{\tan^2 A-\tan^2 B} {1-\tan^2 A\tan^2 B} \] Proof L.H.S. \[ = \frac{\tan(A+B)}{\cot(A-B)} \] \[ = \tan(A+B)\tan(A-B) \] \[ = \frac{\tan A+\tan B}{1-\tan A\tan B} \times \frac{\tan A-\tan B}{1+\tan A\tan B} \] \[ = \frac{(\tan A+\tan B)(\tan A-\tan B)} {(1-\tan

Prove that: cos²A + cos²B − 2 cosA cosB cos(A+B) = sin²(A+B) Question Prove that: \[ \cos^2 A+\cos^2 B-2\cos A\cos B\cos(A+B) = \sin^2(A+B) \] Proof L.H.S. \[ = \cos^2 A+\cos^2 B \] \[ -2\cos A\cos B(\cos A\cos B-\sin A\sin B) \] \[ = \cos^2 A+\cos^2 B-2\cos^2 A\cos^2 B \] \[ +2\sin A\sin B\cos A\cos B