Question

Prove that tan³ theta / 1+tan² theta +cot³ theta/1 + cot² theta
= sec theta cosec theta - 2 sin theta cos theta
​

Answer 1

Step-by-step explanation:

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Answer 2

Answer with Step-by-step explanation:

Given:

LHS

$\frac{tan^3\theta}{1+tan^2\theta}+\frac{cot^3\theta}{1+cot^2\theta}$

We know that

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

$1+cot^2\theta=cosec^2\theta=\frac{1}{sin^2\theta}$

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

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

Using the formula

$\frac{sin^3\theta}{cos^3\theta\times \frac{1}{cos^2\theta}}+\frac{cos^3\theta}{sin^3\theta\times \frac{1}{sin^2\theta}}$

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

$\frac{sin^4\theta+cos^4\theta}{sin\theta cos\theta}$

$\frac{sin^4\theta+cos^4\theta+2sin^2\theta cos^2\theta-2sin^2\theta cos^2\theta}{sin\theta cos\theta}$

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

By using

$(sin^2\theta +cos^2\theta)^2=sin^4\theta+cos^4\theta+2sin^2\theta cos^2\theta$

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

By using the formula

$sin^2\theta+cos^2\theta=1$

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

$sec\theta cosec\theta-2sin\theta cos\theta$

Hence, proved

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