Prove that - sin - cos90- - -cos -sin90- - - - cos - sin90- - -sin -cos90- - - - 1

We have #160;LHS = sin θ cos(90^∘ θ)cos θsin(90^∘ #160; θ) + cos θ sin(90^∘ θ)sin θcos(90^∘ θ) #160; ⇒ LHS = sin θ ...

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Integration by Parts

Prove that : ...

Question

Prove that :
$\frac{s i n θ c o s (90^{\circ} - θ) c o s θ}{s i n (90^{\circ} - θ)} + \frac{c o s θ s i n (90^{\circ} - θ) s i n θ}{c o s (90^{\circ} - θ)} = 1$

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sinθ \cos (90 ° - θ) + \sin (90 ° - θ) cosθ = 1

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\frac{cos θ sin (90^{\circ} - θ) sin θ}{sec (90^{\circ} - θ)}

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\frac{{sin}^{2} 20^{\circ} + {sin}^{2} 70^{\circ}}{{cos}^{2} 20^{\circ} + {cos}^{2} 70^{\circ}} + \frac{sin (90^{\circ} - θ) sin θ}{tan θ} + \frac{cos (90^{\circ} - θ) cos θ}{cot θ}

c o s θ s i n (90^{\circ} - θ) + s i n θ c o s (90^{\circ} - θ) = 1.

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