Question

Prove the expression $({\sec ^2}\theta - 1)(1 - \cos e{c^2}\theta ) = - 1$

Answer 1

Hint: First, we need to analyze the given information which is in the trigonometric form.
$\bullet$ The trigonometric functions are useful whenever the trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
$\bullet$ Since all the six trigonometric functions are related to each other, we will convert the given into some form to simplify the equation.
Formula used:
${\tan ^2}\theta = {\sec ^2}\theta - 1$ because ${\sec ^2}\theta - {\tan ^2}\theta = 1$
$ - {\cot ^2}\theta = 1 - \cos e{c^2}\theta $ because $\cos e{c^2}\theta - {\cot ^2}\theta = 1$

Complete step-by-step solution:
Given that $({\sec ^2}\theta - 1)(1 - \cos e{c^2}\theta )$ and we need to prove that this value has the exact number as $ - 1$ equals. As we said all the trigonometric values are related to each other like, $\sin x = \dfrac{1}{{\sec x}}$ or also like $\tan x = \dfrac{1}{{\cot x}}$
Now we are going to convert the given into some form to simplify easily.
Since we know that ${\tan ^2}\theta = {\sec ^2}\theta - 1$ because ${\sec ^2}\theta - {\tan ^2}\theta = 1$. Then apply it in given we get $({\sec ^2}\theta - 1)(1 - \cos e{c^2}\theta ) \Rightarrow ({\tan ^2}\theta )(1 - \cos e{c^2}\theta )$
Also, we know that $ - {\cot ^2}\theta = 1 - \cos e{c^2}\theta $ because $\cos e{c^2}\theta - {\cot ^2}\theta = 1$ then we have \[({\tan ^2}\theta )( - {\cot ^2}\theta )\]
Thus, applying the value of the cot or tan, then we get $\cot \theta = \dfrac{1}{{\tan \theta }}$
Hence, we have \[({\tan ^2}\theta )( - {\cot ^2}\theta ) \Rightarrow ({\tan ^2}\theta )( - \dfrac{1}{{{{\tan }^2}\theta }})\].
Now canceling the common terms, we get \[({\tan ^2}\theta )( - \dfrac{1}{{{{\tan }^2}\theta }}) = - 1\]
Thus, we proved that $({\sec ^2}\theta - 1)(1 - \cos e{c^2}\theta ) = - 1$ using the trigonometric functions of relations.

Note: In total there are six trigonometric values which are sine, cos, tan, sec, cosec, cot while all the values have been relation like $\dfrac{{\sin }}{{\cos }} = \tan $and $\tan = \dfrac{1}{{\cot }}$
We found the values using the three relations on the trigonometric values are ${\sec ^2}\theta - {\tan ^2}\theta = 1$ is the relation of the secant and tangent. $\cos e{c^2}\theta - {\cot ^2}\theta = 1$ is the relation of the cosecant and cotangent. Finally, we used $\tan = \dfrac{1}{{\cot }}$, or we can able to use the inverse process like $\cot \theta = \dfrac{1}{{\tan \theta }}$
Both will get the same answer like \[({\tan ^2}\theta )( - {\cot ^2}\theta ) \Rightarrow ({\tan ^2}\theta )( - \dfrac{1}{{{{\tan }^2}\theta }})\] will get minus one. Also \[({\tan ^2}\theta )( - {\cot ^2}\theta ) \Rightarrow (\dfrac{1}{{{{\cot }^2}\theta }})( - {\cot ^2}\theta )\] will get the same minus one and hence both the methods are inverse images to each other.