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Composition of Trigonometric Functions and Inverse Trigonometric Functions

Prove that s...

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Prove that sin(50°+θ)-cos(40°-θ)=0

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cos (40^{\circ} + θ) - sin (50^{\circ} - θ) =

s i n (50 + θ) - c o s (40 - θ) = ?

Q. Prove the following :

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

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

tan 20 tan 70 tan 80 tan 89 = 1

cos (40 - θ) - sin (50 + θ) + \frac{{cos}^{2} 40 + {cos}^{2} 50}{{sin}^{2} 40 + {sin}^{2} 50}

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Composition of Trigonometric Functions and Inverse Trigonometric Functions

Standard XII Mathematics

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