Question

if cosecX-sinX =a and secX-cosX=b then 1/ab[1/a+1/b]

Answer 1

hope this wil help u

Answer 2

The formula will use to solve the problem

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

2. $tan x=\frac{sinx}{cosx}$

3. $cotx=\frac{cosx}{sin x}$

Given cosecx-sinx =a and secx-cosx=b.

We have to find the value of $\frac{1}{ab}[ \frac{1}{a}+\frac{1}{b}]$.

First, find the value of a and b

$a=\frac{1}{sinx } -sinx$

  $=\frac{1-sin^{2}x }{sinx}$

  $=\frac{ \cos ^{2}x}{\sin x}$

  $=\cos x\cot x$

and

$b=\frac{1}{\cos x}-\cos x$

  $=\frac{1-\cos ^{2}x}{\cos x}$

  $=\frac{\sin ^{2}x}{\cos x}$

 $=\sin x\tan x$

Now find the value of given expression

$\frac{1}{ab}[ \frac{1}{a}+\frac{1}{b}]=\frac{1}{(\cos x\cot x)(\sin x\tan x)}[ \frac{1}{\cos x\cot x}+\frac{1}{\sin x\tan x}]$

               $=\frac{\sin x\tan x+\cos x\cot x}{(\cos x\cot x\sin x\tan x)^2}$

               $=\frac{\sin x\tan x+\cos x\cot x}{(\cos x\sin x)^2}$

              $=\frac{\sin x\tan x}{(\cos x\sin x)^2}+\frac{\cos x\cot x}{(\cos x\sin x)^2}$

             $=\frac{1}{\sin ^{3}x}+\frac{1}{\cos ^{3}x}$

Hence the value is $\frac{1}{\sin ^{3}x}+\frac{1}{\cos ^{3}x}$.