If cosec θ = 5/3, then evaluate (sec² θ -1) cot² θ (1 + cot² θ )

If cosec θ = 5/3, then evaluate (sec² θ -1) cot² θ (1 +…

If cosec $\theta$ = 5/3, then evaluate (sec² $\theta$ -1) cot² $\theta$ (1 + cot² $\theta$ )

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Correct answer: C

Explanation

The correct answer is Option 3 i.e. $\frac{25}{9}$ .

cosec $\theta$ = 5/3

H = 5 and P = 3

(3, 4, 5) are the Pythagorean triplets

(sec² $\theta$ -1) cot² $\theta$ (1 + cot² $\theta$ )

⇒ ${(({5 \over 4})^2 - 1})({4\over 3})^2({1 + ({4 \over 3})^2})$

⇒ ${({25 \over 16} - 1})({16\over 9})({1 + {16 \over 9}})$

⇒ ${({9 \over 16})}({16\over 9}){({25 \over 9})}$ = $25 \over 9$