Evaluate - 65- - - - 25- - - - tan55- - - - 35- - -

We know, (90 θ)=θ (90 θ)=tanθ (65^∘+θ)=(90^∘ (25^∘ θ))=(25^∘ θ) (35^∘+θ)=tan(90^∘ (55^∘ θ))=tan(55^∘ θ) ∴(65^∘ + θ) (25^∘ θ) ...

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Complementary Trigonometric Ratios

Evaluate : ...

Question

Evaluate :
$csc (65^{\circ} + θ) - sec (25^{\circ} - θ) - tan (55^{\circ} - θ) + cot (35^{\circ} + θ)$

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Prove that:

(i) $s i n (70^{\circ} + θ) - c o s (20^{\circ} - θ) = 0$

(ii) $t a n (55^{\circ} - θ) - c o t (35^{\circ} + θ) = 0$

(iii) $c o s e c (67^{\circ} + θ) - s e c (23^{\circ} - θ) = 0$

(iv) $c o s e c (65^{\circ} + θ) - s e c (25^{\circ} - θ) - t a n (55^{\circ} - θ) + c o t (35^{\circ} + θ) = 0$

(v) $s i n (50^{\circ} + θ) - c o s (40^{\circ} - θ) + t a n 1^{\circ} t a n 10^{\circ} t a n 80^{\circ} t a n 89^{\circ} = 1$

[cosec(75^{\circ} + θ) - sec (15^{\circ} - θ) - tan (55^{\circ} + θ) + cot (35^{\circ} - θ)]

2 ({sin}^{6} θ + {cos}^{6} θ) - 3 ({sin}^{4} θ + {cos}^{4} θ) + 1

c o s e c (75^{\circ} + θ) - s e c (15^{\circ} - θ) - t a n (55^{\circ} + θ) + c o t (35^{\circ} - θ)

\frac{3}{2}

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