Question

Evaluate: (〖cos〗^2 (45°+θ)+〖cos〗^2 (45°-θ))/(tan (60°+θ)tan (30°-θ) )+cosec (75°+θ)-sec⁡(15°-θ).
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Answer 1

Answer:

Taking Left Hand Side,\begin{gathered}\dfrac{cos^2(45+\theta)+cos^2(45-\theta)}{tan(60+\theta).tan(30-\theta)}\\\\\\=\dfrac{cos^(45+\theta)+sin^2(90-45+\theta)}{tan(60+\theta).cot(90-30+\theta)}\\\\\\=\dfrac{cos^2(45+\theta)+sin^2(45+\theta)}{tan(60+\theta).cot(60+\theta)}\\\\\\=\dfrac{1}{1}\\\\\\=1\end{gathered}

tan(60+θ).tan(30−θ)

cos

2

(45+θ)+cos

2

(45−θ)

=

tan(60+θ).cot(90−30+θ)

cos

(

45+θ)+sin

2

(90−45+θ)

=

tan(60+θ).cot(60+θ)

cos

2

(45+θ)+sin

2

(45+θ)

=

1

1

=1

Right Hand Side = Left Hand Side.

Hence Proved.