Evaluate-cosec 90-tan- 90-sin 255-sin 235-tan 10-tan 20-tan 60-tan 70-tan 80-

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Complementary Trigonometric Ratios

Evaluate:θcos...

Question

Evaluate:

$\frac{\sec θ cosec (90 ° - θ) - \tan θ \cot (90 ° - θ) + \sin^{2} 55 ° + \sin^{2} 35 °}{\tan 10 ° \tan 20 ° \tan 60 ° \tan 70 ° \tan 80 °}$

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Solution

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\frac{secθ + tanθ - 1}{tanθ - secθ + 1} = \frac{cosθ}{1 - sinθ}

\frac{secθ cosec (90 ° - θ) - tanθ \cot (90 ° - θ) + \sin^{2} 55 ° + \sin^{2} 35 °}{\tan 10 ° \tan 20 ° \tan 60 ° \tan 70 ° \tan 80 °}

\frac{s e c θ c o s e c (90^{\circ} - θ) - t a n θ c o t (90^{\circ} - θ) + s i n^{2} 55^{\circ} + s i n^{2} 35^{\circ}}{t a n 10^{\circ} t a n 20^{\circ} t a n 60^{\circ} t a n 70^{\circ} t a n 80^{\circ}}

\frac{- tanθ \cdot \cot (90 ° - θ) + secθ cosec (90 ° - θ) + \sin^{2} 35 ° + \sin^{2} 55 °}{\tan 10 ° \tan 20 ° \tan 30 ° \tan 70 ° \tan 80 °}

\frac{\sin^{2} 40 ° + \sin^{2} 50 °}{\cos^{2} 20 ° + \cos^{2} 70 °} + \tan 10 ° \tan 20 ° \tan 60 ° \tan 70 ° \tan 80 °

sin (50^{\circ} + θ) - cos (40^{\circ} - θ) + tan 1^{\circ} tan 10^{\circ} tan 20^{\circ} tan 70^{\circ} tan 80^{\circ} tan 89^{\circ} = 1

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