Question

Prove that (sin 0+ cosec 0)^2 +(cos 0 + sec 0)^2 = 7+tan^2 0+ cot^2 0.

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Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer with Step-by-step explanation:

LHS

$(sin\theta+cosec\theta)^2+(cos\theta+sec\theta)^2$

$sin^2\theta+cosec^2\theta+2sin\theta cosec\theta+sec^2\theta+cos^2\theta+2cos\theta sec\theta$

Using identity :$(a+b)^2=a^2+b^2+2ab$

We  know that

$tan^2\theta+1=sec^2\theta, 1+cot^2\theta=cosec^2\theta,sec\theta=\frac{1}{cos\theta},cosec\theta=\frac{1}{sin\theta}$

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

Using the formula

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

$3+tan^2\theta+cot^2\theta+2+2$

$7+tan^2\theta+cot^2\theta$

LHS=RHS

Hence, proved.

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