Question

Prove that 1 + cos theta minus sin square theta by sin theta into 1 + cos theta is equal to cot theta

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

$\tt{\rightarrow\dfrac{1+cos\theta -sin^{2}\theta}{sin\theta (1+cos\theta)}}$

$\tt{\rightarrow\dfrac{(1-sin^{2}\theta) + cos\theta}{sin\theta (1+cos\theta)}}$

★Identity :-

1 - sin^2θ = cos^2θ

$\tt{\rightarrow\dfrac{(cos^{2}\theta) + cos\theta}{sin\theta (1+cos\theta)}}$

$\tt{\rightarrow\dfrac{cos\theta(1 + cos\theta)}{sin\theta (1+cos\theta)}}$

$\tt{\rightarrow\dfrac{(cos\theta)}{(sin\theta)}}$

= cotθ

Some additional identities :-

★tan θ = sin θ /cos θ

★cot θ = cos θ / sin θ

★(sin² θ) + (cos² θ) = 1

★1 + tan² θ = sec² θ

★1+ cot² θ = cosec² θ

Answer 2

Step-by-step explanation:

(1+cot A - sin^2 A)/sin A(1+cos A )= cot A

LHS = (1+cot A - sin^2 A)/sin A(1+cos A )

=( 1- sin^2 A+cos A)/ sin A(1+cos A )

= (cos^2 A +cos A)/ sin A (1+cos A )

= cos A (cos A +1)/sin A(1+cos A )

= cot A