Question

1+cot theta + tan theta into sin theta - cos theta is equal to sec cube theta minus cosec cube theta upon sec square theta into cosec square theta

Answer 1

Answer with Step-by-step explanation:

LHS:

$(1+cot\theta+tan\theta)(sin\theta-cos\theta)$

We know that

$tan\theta=\frac{sin\theta}{cos\theta}$

$cot\theta=\frac{cos\theta}{sin\theta}$

$(1+\frac{cos\theta}{sin\theta}+\frac{sin\theta}{cos\theta})(sin\theta-cos\theta)$

$(\frac{sin\theta cos\theta+cos^2\theta+sin^2\theta}{sin\thetacos\theta})(sin\theta-cos\theta)$

We know that

$sin^3\theta-cos^3\theta=(sin\theta-cos\theta)(sin^2\theta+cos^2\theta+sin\thetacos\theta)$

$\frac{sin^3\theta-cos^3\theta}{sin\theta cos\theta}$

$\frac{sin^3\theta}{sin\theta cos\theta}-\frac{cos^3\theta}{sin\theta cos\theta}$

$\frac{sin^2\theta}{cos\theta}-\frac{cos^2\theta}{sin\theta}$

RHS:

$\frac{sec^3\theta-cosec^3\theta}{cosec^2\theta sec^2\theta}$

$\frac{sec\theta}{cosec^2\theta}-\frac{cosec\theta}{sec^2\theta}$

$\frac{sin^2\theta}{cos\theta}-\frac{cos^2\theta}{sin\theta}$

Using the formula

$sec\theta=\frac{1}{cos\theta}, cosec\theta=\frac{1}{sin\theta}$

LHS=RHS

Hence, proved.

#Learns more:

https://brainly.in/question/15636659:Answered by Nishatalwade