Question

Show that ${\cos ^4}\alpha + 2{\cos ^2}\alpha (1 - \dfrac{1}{{{{\sec }^2}\alpha }}) = 1 - {\sin ^4}\alpha$

Answer 1

To solve such questions, we have to simplify LHS (Left Hand Side) and try to simplify it and convert it into RHS (Right Hand Side). We use trigonometric identities such as:${\sin ^2}\alpha + {\cos ^2}\alpha = 1$ and $(a - b)(a + b) = {a^2} - {b^2}$.

Complete step by step answer:
Let’s start by writing down the left hand side:
${\cos ^4}\alpha + 2{\cos ^2}\alpha (1 - \dfrac{1}{{{{\sec }^2}\alpha }})$
We know that $\dfrac{1}{{{{\sec }^2}\alpha }} = {\cos ^2}\alpha$
${\cos ^4}\alpha + 2{\cos ^2}\alpha (1 - {\cos ^2}\alpha )$
After taking out ${\cos ^2}\alpha$ common from the equation, we get:
${\cos ^2}\alpha [{\cos ^2}\alpha + 2(1 - {\cos ^2}\alpha )]$
We know that ${\cos ^2}\alpha = 1 - {\sin ^2}\alpha$

After simplifying the equation in the bracket we get:
${\cos ^2}\alpha + 2(1 - {\cos ^2}\alpha )$
$\Rightarrow {\cos ^2}\alpha + 2 - 2{\cos ^2}\alpha$
$\Rightarrow 2 - {\cos ^2}\alpha$
$\Rightarrow 1 + 1 - {\cos ^2}\alpha$
$\Rightarrow 1 + {\sin ^2}\alpha$
Hence after multiplying the two terms, we get:
$(1 - {\sin ^2}\alpha )(1 + {\sin ^2}\alpha ) = 1 - {\sin ^4}\alpha$
Since LHS = RHS, hence proved.

Note: This question tests the fundamentals of trigonometric identities, hence remembering them will reduce the difficulty of the question and complexity. When we simplify the left hand side, we have to simplify it in such a way that it is similar to the equation on the right hand side. Sometimes going from left to right might not help and be difficult, then we can use the right to left approach and simplify the equation.