Question

Evaluate: $[\frac{1}{{\sec }^{2}x-{\cos }^{2}x}+\frac{1}{cose{c}^{2}x-{\sin }^{2}x}]\times {\sin }^{2}x\cdot {\cos }^{2}x.$

• A. $\frac{1-{\sin }^{2}x\cdot {\cos }^{2}x}{2-{\sin }^{2}x\cdot {\cos }^{2}x}$.
• B. $\frac{1-{\sin }^{2}x\cdot {\cos }^{2}x}{2+{\sin }^{2}x\cdot {\cos }^{2}x}$.
• C. $\frac{1+{\sin }^{2}x\cdot {\cos }^{2}x}{2+{\sin }^{2}x\cdot {\cos }^{2}x}$.
• D. $\frac{1-{\sin }^{2}x\cdot {\cos }^{2}x}{2+sinxcosx}$.

Answer 1

The correct option is B $\frac{1-{\sin }^{2}x\cdot {\cos }^{2}x}{2+{\sin }^{2}x\cdot {\cos }^{2}x}$.

We are given, $[\frac{1}{{\sec }^{2}x-{\cos }^{2}x}+\frac{1}{{\csc }^{2}x-{\sin }^{2}x}]\times {\sin }^{2}x.{\cos }^{2}x$

We know thta, $\cos \theta .\sec \theta =1\sin \theta =1$ & $\sin \theta .\csc \theta =1$

So, $\sec x=\frac{1}{\cos x}$ & $\csc x=\frac{1}{\sin x}$

$\Rightarrow [\frac{1}{{\sec }^{2}x-{\cos }^{2}x}+\frac{1}{{\csc }^{2}x-{\sin }^{2}x}]\times {\sin }^{2}x.{\cos }^{2}x$

$=[\frac{1}{\frac{1}{{\cos }^{2}x}-{\cos }^{2}x}+\frac{1}{\frac{1}{{\sin }^{2}x}-{\sin }^{2}x}]\times {\sin }^{2}x.{\cos }^{2}x$

$=[\frac{{\cos }^{2}x}{(1-{\cos }^{4}x)}+\frac{{\sin }^{2}x}{(1-{\sin }^{4}x)}]\times {\sin }^{2}x.{\cos }^{2}x$

$=\left[\frac{{\cos }^{2}a(1-{\sin }^{4}x)+{\sin }^{2}x(1-{\cos }^{2}x)}{(1-{\cos }^{4}x)(1-{\sin }^{4}x)}\right]\times {\sin }^{2}x.{\cos }^{2}x$

$=\left[\frac{{\cos }^{2}x-{\cos }^{2}x{\sin }^{4}x+{\sin }^{2}x-{\sin }^{2}x{\cos }^{4}x}{(1-{\cos }^{2}x)(1+{\cos }^{2}x)(1+{\sin }^{2}x)}\right]\times {\sin }^{2}x.{\cos }^{2}x$

(now, $(a^2-b^2)=(a-b)(a+b)$ & ${\sin }^2\theta +{\cos }^2\theta =1$)

So, $=\frac{[1-{\cos }^{2}x{\sin }^{2}x({\sin }^{2}x+{\cos }^{2}x\left)\right]}{\left({\sin }^{2}x\right)(1+{\cos }^{2}x)\left({\cos }^x\right)(1+{\sin }^{2}x)}\times {\sin }^{2}x{\cos }^{2}x$

$=\frac{[1-{\cos }^{2}x{\sin }^{2}x]}{(1+{\cos }^{2}x)(1+{\sin }^{2}x)}$

$=\frac{1-{\cos }^{2}x{\sin }^{2}x}{1+({\sin }^{2}x+{\cos }^{2}x)+{\sin }^{2}x{\cos }^{2}x}$

$=\frac{1-{\sin }^{2}x.{\cos }^{2}x}{1+1+{\sin }^{2}x.{\cos }^{2}x}$

$=\frac{1-{\sin }^{2}x.{\cos }^{2}x}{2+{\sin }^{2}x.{\cos }^{2}x}$