Sin50-cos40-tan 1-tan 10- tan 20- tan 70- tan 80- tan 89-

Sin50 #176;+ #952; cos40 #176; #952;+tan1 #176; #160;tan10 #176; #160;tan20 #176; #160;tan70 #176; #160;tan80 #176; #160;tan89 #1 ...

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Trigonometric Ratios of Compound Angles

sin50∘+θ-cos4...

Question

sin (50° + θ) – cos (40° – θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89°

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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

\frac{\cos (90 ° - θ) \sec (90 ° - θ) \tan θ}{cosec (90 ° - θ) \sin (90 ° - θ) \cot (90 ° - θ)} + \frac{\tan (90 ° - θ)}{\cot θ} = 2

\frac{\tan (90 ° - A) \cot A}{{cosec}^{2} A} - \cos^{2} A = 0

\frac{\cos (90 ° - A) \sin (90 ° - A)}{\tan (90 ° - A)} = \sin^{2} A

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

sin (50 + θ) - cos (40 - θ) + tan 1 tan 10

tan 20 tan 70 tan 80 tan 89 = 1

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