Question

Solve this question
I got tan theta +sec theta in LHS side

Answer 1

divide throughout by cos theta as we will be needing to prove it that way...

when you do so :

tan theta + sec theta - 1/tan theta - sec theta + 1

we know that sec2 theta - tan2 theta is 1 so we put that in the place of 1

tan theta + sec theta - ( sec2 theta - tan2 theta) / tan theta - sec theta + 1

a2+b2 = (a+b) (a-b)

tan theta + sec theta - (sec theta + tan theta) ( sec theta - tan theta)/tan theta - sec theta + 1

we take tan theta + sec theta as common in the numerator

tan theta + sec theta * ( tan theta - sec theta +1 )/ tan theta - sec theta +1

tan theta - sec theta +1  gets cancelled and we get tan theta + sec theta

now coming to the R.H.S

rationalizing :

sec theta + tan theta/sec2 theta - tan2 theta

sec2 theta - tan2 theta is 1

so it becomes sec theta + tan theta

so L.H.S =R.H.S

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Answer 2

$\bold{\underline{Question:}}$

$\bold{\frac{Sin\theta-Cos\theta+1}{Sin\theta+Cos\theta-1}=\frac{1}{Sec\theta-tan\theta} }$

$\bold{\underline{Solution:}}$

From L.H.S.

$\bold{=\frac{Sin\theta-Cos\theta+1}{Sin\theta+Cos\theta-1}}$

$\bold{\underline{\textsf{Dividing numerator and denominator by}}}\quad{\bold{Cos\theta}}.$

$\bold{We\:get,}$

$\bold{=\frac{\frac{Sin\theta}{Cos\theta}-\frac{Cos\theta}{Cos\theta}+\frac{1}{Cos\theta}}{\frac{Sin\theta}{Cos\theta}+\frac{Cos\theta}{Cos\theta}-\frac{1}{Cos\theta}}}$

$\bold{=\frac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}}$

$\bold{\boxed{Note:sec^2\theta-tan^2\theta=1(Identity)}}$

$\underline{\bold{\textsf{Putting the value of 1 in numerator we get,}}}$

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

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

$\bold{=tan\theta+sec\theta.}$

$\bold{\boxed{Sec\theta+tan\theta=\frac{1}{sec\theta-tan\theta}.}}$

Hence proved.