Question

Evaluate: (〖cos〗^2 (45°+θ)+〖cos〗^2 (45°-θ))/(tan (60°+θ)tan (30°-θ) )+cosec (75°+θ)-sec⁡(15°-θ).​

Answer 1

Answer:-

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$\implies{\dfrac{cos^2 (45 \degree+ \theta)+ cos^2 (45 \degree - \theta)}{tan(60 \degree + \theta) tan(30 \degree - \theta)} + cosec (75 \degree + \theta)- sec(15 \degree - \theta)}$

$\implies{\dfrac{cos^2 (45 \degree+ \theta)+ sin^2 (90 \degree -(45 \degree - \theta))}{tan(60 \degree + \theta)cot (90 \degree - (30 \degree - \theta))} + cosec (75 \degree + \theta)- cosec(90 \degree -(75 \degree + \theta))}$

$\implies{\dfrac{cos^2 (45 \degree+ \theta)+ sin^2 (45 \degree + \theta)}{tan(60 \degree + \theta)cot(60 \degree + \theta))} + cosec (75 \degree + \theta)- cosec(75 \degree + \theta))}$

$\implies{\dfrac{cos^2 (45 \degree+ \theta)+ sin^2 (45 \degree + \theta)}{tan(60 \degree + \theta)cot(60 \degree + \theta))}\cancel{ + cosec (75 \degree + \theta)}\cancel {- cosec(75 \degree + \theta))}}$

$\implies{\dfrac{cos^2 (45 \degree+ \theta)+ sin^2 (45 \degree + \theta)}{tan(60 \degree + \theta)cot(60 \degree + \theta))}}$

We know,

$\boxed{sin^2 \theta + cos^2 \theta = 1}$

Also,

$\boxed{tan \theta \times \cot \theta = 1}$

Therefore,

$\implies\dfrac{1}{1} + 0$

$\implies\bf{1}$

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