Question

Prove that :
$\sf{\dfrac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}= sec\theta+tan\theta=\dfrac{1+sin\theta}{cos\theta}$

General Instructions :
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Use G00GLE for answering but answer should be correct .

Answer 1

Answer:

Step-by-step explanation:

To Prove :-

$\dfrac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}=sec\theta+tan\theta=\dfrac{1+sin\theta}{cos\theta}$

Solution :-

Taking L.H.S :-

$= \dfrac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}$

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

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

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

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

$=sec\theta+tan\theta$

$=\dfrac{1}{cos\theta}+\dfrac{sin\theta}{cos\theta}$

$=\dfrac{1+sin\theta}{cos\theta}$

= R.H.S