Question

tennetiraj bhaiya

i said solve this with change into sin and cos only

plzz solve with sin and cos only ​

Answer 1

Answer:

answer for the given problem is given changing them into sin and cos

Answer 2

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

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

Now ,Taking L.H.S and rationalising it

$= > \frac{sec \theta + tan \theta}{sec \theta - tan \theta} \times \frac{sec \theta + tan \theta}{sec \theta + tan \theta}$

Identity used here :-

$\bold{\boxed{{(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

$\bold{\boxed{{x}^{2} - {y}^{2} = (x + y)(x - y)}}$

$= > \frac{ {(sec \theta + tan \theta )}^{2} }{ {sec}^{2} \theta - {tan}^{2} \theta } = {(sec \alpha + tan \alpha )}^{2}$✓

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

$\bold{\boxed{As\: ( {sec}^{2} \theta= 1 + {tan}^{2} \theta )}}$

It can be further solve :

$= > {(sec \theta + tan \theta )}^{2} = {sec}^{2} \alpha + {tan}^{2} \theta + 2sec \theta\tan \theta$

$= 1 + {tan}^{2} \theta + {tan}^{2} \theta + 2sec \theta tan \theta$

$= 1 + 2 {tan}^{2} \theta + 2sec \theta\tan \theta$