Question

Show that $(1+{\tan }^2\theta )(1-\sin \theta )(1+\sin \theta )=1$

Answer 1

Let usfirstfind the value of left hand side (LHS) that is $(1+{\tan }^2\theta )(1-\sin \theta )(1+\sin \theta )$ as shown below:

$(1+{\tan }^2\theta )(1-\sin \theta )(1+\sin \theta )=(1+{\tan }^2\theta )(1-{\sin }^2\theta )(\because (a-b\left)\right(a+b)=a^2-b^2)={\sec }^2\theta \times {\cos }^2\theta (\because {\sec }^{2}x=1+{\tan }^{2}x,1-{\sin }^2\theta ={\cos }^2\theta )=\frac{1}{{\cos }^2\theta }\times {\cos }^2\theta (\because \sec x=\frac{1}{\cos x})=1=RHS$

Since LHS=RHS,

Hence, $(1+{\tan }^2\theta )(1-\sin \theta )(1+\sin \theta )=1$.