Question

Hey guys, please help me prove this identity by making one side equal to the other. ​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

${cos}^{2} \theta = 1 - {sin}^{2} \theta$

TO PROVE

$\displaystyle \: \frac{cos \theta \: }{tan\theta \:(1 - sin \: \theta )\:} = 1 + \frac{1}{sin \: \theta \:}$

PROOF

$\displaystyle \: \frac{cos \theta \: }{tan\theta \:(1 - sin \: \theta )\:}$

$= \displaystyle \: \frac{cos \theta \: }{ \frac{sin \theta}{cos\theta} \:(1 - sin \: \theta )\:}$

$= \displaystyle \: \frac{{cos}^{2} \theta \: }{sin\theta \:(1 - sin \: \theta )\:}$

$= \displaystyle \: \frac{1 - {sin}^{2} \theta \: }{sin\theta \:(1 - sin \: \theta )\:}$

$= \displaystyle \: \frac{(1 + sin \: \theta ) (1 - sin \: \theta )}{sin\theta \:(1 - sin \: \theta )\:}$

$= \displaystyle \: \frac{(1 + sin \: \theta ) }{sin\theta \:\:}$

$\displaystyle \: = \frac{1}{sin \: \theta \:} + 1$

$\displaystyle \: = 1 + \frac{1}{sin \: \theta \:}$

Hence proved

Answer 2

Answer:

=

cosθ

sinθ

(1−sinθ)

cosθ

= \displaystyle \: \frac{{cos}^{2} \theta \: }{sin\theta \:(1 - sin \: \theta )\:}=

sinθ(1−sinθ)

cos

2

θ

= \displaystyle \: \frac{1 - {sin}^{2} \theta \: }{sin\theta \:(1 - sin \: \theta )\:}=

sinθ(1−sinθ)

1−sin

2

θ

= \displaystyle \: \frac{(1 + sin \: \theta ) (1 - sin \: \theta )}{sin\theta \:(1 - sin \: \theta )\:}=

sinθ(1−sinθ)

(1+sinθ)(1−sinθ)

= \displaystyle \: \frac{(1 + sin \: \theta ) }{sin\theta \:\:}=

sinθ

(1+sinθ)

\displaystyle \: = \frac{1}{sin \: \theta \:} + 1=

sinθ

1

+1

\displaystyle \: = 1 + \frac{1}{sin \: \theta \:}=1+

sinθ

1