If sinθ=x2−y2x2+y2, then find the values of cosθ and 1tanθ
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Solution
we know that sinθ=sideoppositetoangleθhypotenuseorsinθ=Perpendicularhypotenuse sinθ=x2−y2x2+y2⇒PH=x2−y2x2+y2⇒ABBC=x2−y2x2+y2 side opposite to angle θ=AB=x2−y2 hypotenuse AC=x2+y2 In right angled △ABC, we have ⇒(AB)2+(BC)2=(AC)2⇒(x2−y2)2+(BC)2=(x2+y2)2⇒(BC)2=(x2+y2)2−(x2−y2)2 (By Pythagoras theorem) ⇒(BC)2=[x2+y2+x2−y2][x2+y2−(x2−y2)]=(2x2)(2y2) (using identity a2−b2=(a+b)(a−b)) BC=√4x2y2=±2xy taking positie square root since, side cannot be negative cosθ=Basehypotenuse=BCAC=2xyx2+y2 and tanθ=PerpendicularBase=ABBC=x2−y22xy so,
1tanθ=1x2−y22xy=2xyx2−y2