1+SinA.1−SinA=\sqrt{1+Sin A}.\sqrt{1-Sin A} =1+SinA.1−SinA=
Sin θ\thetaθ = ab\frac{a}{b}ba ; Cos θ\thetaθ = cd\frac{c}{d}dc then Cot θ\thetaθ =
Sin θ\thetaθ . Cot θ\thetaθ . Sec θ\thetaθ =
1+Tan2θ1+Cot2θ=\frac{\sqrt{1+Tan^2 \theta}}{\sqrt{1+Cot^2 \theta}} =1+Cot2θ1+Tan2θ=
Sin3θ Cosθ+Cos3θ Sinθ=Sin^3 \theta \, Cos \theta + Cos^3 \theta \, Sin \theta =Sin3θCosθ+Cos3θSinθ=
Tan θ\thetaθ + Cot θ\thetaθ = 2, then Tan2θ+Cot2θTan^2 \theta + Cot^2 \thetaTan2θ+Cot2θ =
1+Sinθ1−Sin2θ=\frac{1+Sin \theta}{\sqrt{1-Sin^2 \theta}} =1−Sin2θ1+Sinθ=
Sin αCos α.Cot α=\frac{Sin\, \alpha}{Cos\, \alpha} . Cot\, \alpha =CosαSinα.Cotα=
If A, B and C are interior angles of a triangle ABC, then Sin (B+C2\frac{B + C}{2}2B+C ) =
1−Tan245∘1+Tan245∘\frac{1-Tan^2 45^\circ}{1+Tan^2 45^\circ}1+Tan245∘1−Tan245∘ can not be expressed as ………