If \[ a=\frac{2\sin x}{1+\cos x+\sin x} \] Prove that \[ \frac{1-\cos x+\sin x}{1+\sin x}=a \] Solution: \[ a=\frac{2\sin x}{1+\cos x+\sin x} \] Multiply numerator and denominator by \[ 1-\cos x+\sin x \] \[ a= \frac{2\sin x(1-\cos x+\sin x)} {(1+\cos x+\sin x)(1-\cos x+\sin x)} \] \[ = \frac{2\sin x(1-\cos x+\sin x)} {(1+\sin x)^2-\cos^2 x} \] \[
Prove the Identity : \[ \cos x(\tan x+2)(2\tan x+1) = 2\sec x+5\sin x \] Solution: \[ \cos x(\tan x+2)(2\tan x+1) \] \[ = \cos x(2\tan^2 x+5\tan x+2) \] \[ = 2\cos x\tan^2 x + 5\cos x\tan x + 2\cos x \] \[ = 2\cos x\cdot\frac{\sin^2 x}{\cos^2 x} + 5\sin x + 2\cos x \] \[
Prove the following identity : cos x (tan x + 2) (2 tan x + 1) = 2 sec x + 5 sin x Read More »
Prove the Identity : \[ \frac{2\sin x\cos x-\cos x} {1-\sin x+\sin^2 x-\cos^2 x} = \cot x \] Solution: \[ \frac{\cos x(2\sin x-1)} {1-\sin x+\sin^2 x-\cos^2 x} \] Using \[ \cos^2 x=1-\sin^2 x \] \[ = \frac{\cos x(2\sin x-1)} {1-\sin x+\sin^2 x-(1-\sin^2 x)} \] \[ = \frac{\cos x(2\sin x-1)} {2\sin^2 x-\sin x} \] \[ = \frac{\cos
Prove the Identity : \[ \frac{(1+\cot x+\tan x)(\sin x-\cos x)} {\sec^3 x-\cosec^3 x} = \sin^2 x\cos^2 x \] Solution: \[ \frac{\left(1+\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}\right) (\sin x-\cos x)} {\frac{1}{\cos^3 x}-\frac{1}{\sin^3 x}} \] \[ = \frac{ \left( \frac{\sin x\cos x+\cos^2 x+\sin^2 x} {\sin x\cos x} \right) (\sin x-\cos x) } { \frac{\sin^3 x-\cos^3 x} {\sin^3 x\cos^3
Prove the Identity : \[ (1+\tan\alpha\tan\beta)^2 + (\tan\alpha-\tan\beta)^2 = \sec^2\alpha\sec^2\beta \] Solution: \[ (1+\tan\alpha\tan\beta)^2 + (\tan\alpha-\tan\beta)^2 \] \[ = 1+\tan^2\alpha\tan^2\beta +2\tan\alpha\tan\beta \] \[ + \tan^2\alpha+\tan^2\beta -2\tan\alpha\tan\beta \] \[ = 1+\tan^2\alpha+\tan^2\beta +\tan^2\alpha\tan^2\beta \] \[ = (1+\tan^2\alpha)(1+\tan^2\beta) \] \[ = \sec^2\alpha\sec^2\beta \] Hence proved. Next Question / Full Exercise
Prove the following identity : (1 + tan α tan β)^2 + (tan α – tan β)^2 = sec^2 α sec^2 β Read More »
Prove the Identity : \[ \left( \frac{1}{\sec^2 x-\cos^2 x} + \frac{1}{\cosec^2 x-\sin^2 x} \right) \sin^2 x\cos^2 x = \frac{1-\sin^2 x\cos^2 x} {2+\sin^2 x\cos^2 x} \] Solution: \[ \left( \frac{1}{\frac{1}{\cos^2 x}-\cos^2 x} + \frac{1}{\frac{1}{\sin^2 x}-\sin^2 x} \right) \sin^2 x\cos^2 x \] \[ = \left( \frac{\cos^2 x}{1-\cos^4 x} + \frac{\sin^2 x}{1-\sin^4 x} \right) \sin^2 x\cos^2 x \]
Prove the Identity : \[ 1-\frac{\sin^2 x}{1+\cot x}-\frac{\cos^2 x}{1+\tan x} = \sin x\cos x \] Solution: \[ 1-\frac{\sin^2 x}{1+\frac{\cos x}{\sin x}} -\frac{\cos^2 x}{1+\frac{\sin x}{\cos x}} \] \[ = 1-\frac{\sin^3 x}{\sin x+\cos x} -\frac{\cos^3 x}{\sin x+\cos x} \] \[ = 1-\frac{\sin^3 x+\cos^3 x}{\sin x+\cos x} \] \[ = 1- \frac{(\sin x+\cos x)(\sin^2 x-\sin x\cos x+\cos^2 x)}
Prove the Identity : \[ \frac{\tan^3 x}{1+\tan^2 x} + \frac{\cot^3 x}{1+\cot^2 x} = \frac{1-2\sin^2 x\cos^2 x}{\sin x\cos x} \] Solution: \[ \frac{\tan^3 x}{1+\tan^2 x} + \frac{\cot^3 x}{1+\cot^2 x} \] Using \[ 1+\tan^2 x=\sec^2 x \quad \text{and} \quad 1+\cot^2 x=\cosec^2 x \] \[ = \frac{\tan^3 x}{\sec^2 x} + \frac{\cot^3 x}{\cosec^2 x} \] \[ = \frac{\sin^3 x}{\cos
Prove the Identity : \[ \frac{\cos x}{1-\sin x} = \frac{1+\cos x+\sin x}{1+\cos x-\sin x} \] Solution: \[ \frac{\cos x}{1-\sin x} \] \[ = \frac{\cos x(1+\sin x)}{(1-\sin x)(1+\sin x)} \] \[ = \frac{\cos x(1+\sin x)}{1-\sin^2 x} \] \[ = \frac{\cos x(1+\sin x)}{\cos^2 x} \] \[ = \frac{1+\sin x}{\cos x} \] \[ = \frac{(1+\sin x)(1+\cos x)} {\cos
Prove the Identity : \[ (\sec x\sec y+\tan x\tan y)^2 – (\sec x\tan y+\tan x\sec y)^2 =1 \] Solution: \[ =(\sec x\sec y+\tan x\tan y+\sec x\tan y+\tan x\sec y) \] \[ \times (\sec x\sec y+\tan x\tan y-\sec x\tan y-\tan x\sec y) \] \[ = (\sec x+\tan x)(\sec y+\tan y) \] \[ \times (\sec x-\tan x)(\sec