Question

How do you prove$\dfrac{{\sec x}}{{\cot x + \tan x}} = \sin x$?

Answer 1

Hint: To prove the given trigonometric function apply trigonometric identity formulas i.e., applying the double angle formula of cos, as we know that sec, csc and cot are derived from primary functions of sin, cos and tan, hence using these functions we can prove that LHS = RHS of the given function. Here we use the double angle formulas to solve the problem.

Complete step-by-step solution:
The given function is
\[\Rightarrow \sec \left( {2x} \right) = \dfrac{{{{\sec }^2}x}}{{2 - {{\sec }^2}x}}\]
Apply double angle formula of cos
\[\Rightarrow \cos \left( {2A} \right) = \cos A \cdot \cos A - \sin A \cdot \sin A\]
\[\Rightarrow \cos \left( {2A} \right) = {\cos ^2}A - {\sin ^2}A\]
\[\Rightarrow \cos \left( {2A} \right) = 2{\cos ^2}A - 1\]
As we know the formula of \[{\cos ^2}A = 1 - 2{\sin ^2}A\]
\[\Rightarrow \cos \left( {2A} \right) = 1 - 2{\sin ^2}A\]
Applying this formula to the given function we get
\[\Rightarrow \sec \left( {2x} \right) = \dfrac{1}{{\cos \left( {2x} \right)}}\]
\[\Rightarrow \sec \left( {2x} \right) = \dfrac{1}{{2{{\cos }^2}x - 1}}\]
Divide numerator and denominator by \[{\cos ^2}x\] as
\[\Rightarrow \sec \left( {2x} \right) = \dfrac{{{{\sec }^2}x}}{{2 - {{\sec }^2}x}}\]

Thus we have proved that L.H.S = R.H.S

Note: The key point to evaluate any trigonometric function is that we must know all the basic trigonometric functions and their relation and to prove the above function we must know all the basic trigonometric identities with respect to double angle formula of cos and its relation with sine function. Here are some of double angle formula for cos function:
\[\cos \left( {2A} \right) = \cos A \cdot \cos A - \sin A \cdot \sin A\]
\[\cos \left( {2A} \right) = {\cos ^2}A - {\sin ^2}A\]
\[\cos \left( {2A} \right) = 2{\cos ^2}A - 1\]