Chapter 14.1, Problem 50PPSE

Solution

To verify: The trigonometric identity $\frac{1-\sin \theta }{\cos \theta }=\frac{\cos \theta }{1+\sin \theta }$

The trigonometric identity $\frac{1-\sin \theta }{\cos \theta }=\frac{\cos \theta }{1+\sin \theta }$ is verified.

Given information:

The trigonimetric identity is $\frac{1-\sin \theta }{\cos \theta }=\frac{\cos \theta }{1+\sin \theta }$

Concept and Formula Used:

We use the trigonometric identity $x^2-y^2=\left(x+y\right)\left(x-y\right),{\sin }^2\theta +{\cos }^2\theta =1$

Calculation:

The given identity is $\frac{1-\sin \theta }{\cos \theta }=\frac{\cos \theta }{1+\sin \theta }$

Taking RHS, substitute the values and simplify

  $\begin{array}{l}\frac{\cos \theta }{1+\sin \theta }=\frac{\cos \theta }{1+\sin \theta }\times \frac{1-\sin \theta }{1-\sin \theta }\left(\text{Rationalise the denominator}\right)\\ \frac{\cos \theta }{1+\sin \theta }=\frac{\cos \theta \left(1-\sin \theta \right)}{\left(1+\sin \theta \right)\left(1-\sin \theta \right)}\\ \frac{\cos \theta }{1+\sin \theta }=\frac{\cos \theta \left(1-\sin \theta \right)}{\left(1^2-{\sin }^2\theta \right)}\left(x^2-y^2=\left(x+y\right)\left(x-y\right)\right)\\ \frac{\cos \theta }{1+\sin \theta }=\frac{\cos \theta \left(1-\sin \theta \right)}{\left({\cos }^2\theta \right)}\left(1-{\sin }^2\theta ={\cos }^2\theta \right)\\ \frac{\cos \theta }{1+\sin \theta }=\frac{\left(1-\sin \theta \right)}{\cos \theta }\\ =LHS\end{array}$

Hence verified

Conclusion:

The trigonometric identity $\frac{1-\sin \theta }{\cos \theta }=\frac{\cos \theta }{1+\sin \theta }$ is verified.