Question

to prove that :---- - cos theta + sin theta +1 / cos theta + sin theya -1 = sec theta + tan theta​

Answer 1

Answer:

Step-by-step explanation:

Correct Question :-

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

Solution :-

Taking L.H.S :-

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

Dividing Numerator and Denominator with "costheta"

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

We know that :-

1) sinA/cosA = tanA

2) 1/cosA = secA

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

We know that :-

$1) sec^2A-tan^2A = 1$

$\implies \dfrac{sec^2\theta-tan^2\theta+tan\theta+sec\theta}{1-tan\theta+sec\theta}$

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

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

$= sec\theta+tan\theta$