Question

guys Is this solution correct or not....

if not then correct it....
please it's too urgent...

Answer 1

Here is your answer....
Hope this helps you....

Answer 2

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$\mathbb{\underline{\blue{SOLUTION:}}}$

^^ After your answer in attachment

Contd.
taking theta as x

=$\frac{tan x + sec x - (sec^{2} x - tan^{2} x)}{tan x - sec x + 1}$

= $\frac{tan x + sec x - ((sec x + tan x)(sec x - tan x))}{tan x - sec x + 1}$

= $\frac{tan x + sec x - (tan x + sec x)(sec x - tan x)}{tan x - sec x + 1}$

= $\frac{(tan x + sec)[1 -(sec x - tan x)]}{tan x - sec x + 1}$

= $\frac{(tan x + sec)[1 -(sec x - tan x)]}{tan x - sec x + 1}$

= $\frac{(tan x + sec)[1 -sec x + tan x)]}{tan x - sec x + 1}$

= sec x + tan x

Rationalise
= $\frac{(sec x + tan x)(sec x - tan x)}{sec x - tan x}$

= $\frac{sec^{2} x - tan^{2}x}{sec x - tan x}$

= $\frac{1}{sec x - tan x}$

= R.H.S

$\underline{\underline{Hence\: Proved}}$

¶¶¶ Identities Used :
•$(a - b)^{2} = (a+b)(a-b)$
[polynomial Identity]
•$sec^{2} x - tan^{2} x = 1$
[Trigonometric Identity]

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