Question

To prove that ${{\sec }^{2}}\theta -{{\cos }^{2}}\varnothing ={{\sin }^{2}}\varnothing +{{\tan }^{2}}\theta$.

Answer 1

Hint : The expression can be solved by converting the standard trigonometric identities to get the desired result.
Formulas used:
The standard function are as follows:

& {{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \\\ & {{\cos }^{2}}\varnothing =1-{{\sin }^{2}}\varnothing \\\ \end{aligned}$$ **_Complete step-by-step answer_** : First step is to consider the Left Hand side. The given expression can be converted by using the formula. The numerical value 1 is subtracted and the result is obtained. The Left hand side and the Right hand side are equal and hence proved. $$\begin{aligned} & L.H.S={{\sec }^{2}}\theta -{{\cos }^{2}}\varnothing \\\ & L.H.S=1+{{\tan }^{2}}\theta -\left( 1-{{\sin }^{2}}\varnothing \right) \\\ & L.H.S=1+{{\tan }^{2}}\theta -1+{{\sin }^{2}}\varnothing \\\ & L.H.S={{\sin }^{2}}\varnothing +{{\tan }^{2}}\theta \\\ & \\\ & R.H.S={{\sin }^{2}}\varnothing +{{\tan }^{2}}\theta \\\ & \therefore L.H.S=R.H.S\text{ }\left( \text{proved} \right) \\\ \end{aligned}$$ Thus, the given expression is proved. **Note** : The trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables where both sides of the equality are defined. The standard conversion are as follows: $$\begin{aligned} & {{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \\\ & {{\cos }^{2}}\varnothing =1-{{\sin }^{2}}\varnothing \\\ \end{aligned}$$