Question

How do you verify the identity $\sin \theta \tan \theta + \cos \theta = \sec \theta$ ?

Answer 1

Hint : Here, you are asked to verify an identity which includes the trigonometric ratios $\sin \theta$ , $\cos \theta$ , $\tan \theta$ and $\sec \theta$ . What you need to do here is to use all the material or precisely trigonometric properties and identities which involves the trigonometric ratios $\sin \theta$ , $\cos \theta$ , $\tan \theta$ and $\sec \theta$ . First you consider the left-hand side and try to simplify it. By simplifying what is meant is that you try to bring the left-hand side equal to the right-hand side.

Complete step by step solution:
So, first let us consider the left-hand side, in that too first consider the first term, $\sin \theta \tan \theta$. Let us keep the sine of $\theta$ as it is and manipulate tangents of $\theta$ . The tangent of any angle can be written as sine of that angle divided by the cosine of that angle, mathematically, we have
$\tan A = \dfrac{{\sin A}}{{\cos A}}$ .
Take $A = \theta$ and substitute in the above identity of tangent, we get
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ .
Now, substitute this value of tangent of $\theta$ in the first term. We will get
$\sin \theta \tan \theta = \sin \theta \left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right) \\\ \sin \theta \tan \theta = \dfrac{{{{\sin }^2}\theta }}{{\cos \theta }} \;$
So, we have the left-hand side as
$\dfrac{{{{\sin }^2}\theta }}{{\cos \theta }} + \cos \theta$ . We need to simplify this further.
Two fractions can be added together if the denominator of both the fractions are equal. So, we multiply and divide the second term $\cos \theta$ by $\cos \theta$ . We get $\cos \theta = \cos \theta \left( {\dfrac{{\cos \theta }}{{\cos \theta }}} \right) = \dfrac{{{{\cos }^2}\theta }}{{\cos \theta }}$ . You can see that both the terms have a denominator as $\cos \theta$ and hence can be added.

$$\Rightarrow \dfrac{{{{\sin }^2}\theta }}{{\cos \theta }} + \cos \theta = \dfrac{{{{\sin }^2}\theta }}{{\cos \theta }} + \dfrac{{{{\cos }^2}\theta }}{{\cos \theta }} \\\ \Rightarrow \dfrac{{{{\sin }^2}\theta }}{{\cos \theta }} + \cos \theta = \dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\cos \theta }} \;$$

Now, the sum of squares of sine and cosine of an angle is equal to one, mathematically, we have ${\sin ^2}\theta + {\cos ^2}\theta = 1$. We substitute this value in the above simplified version of the left-hand side.
$\Rightarrow \sin \theta \tan \theta + \cos \theta = \dfrac{1}{{\cos \theta }}$ .
Also, the inverse of cosine of any angle is equal to secant of that angle and hence, finally we get
$\sin \theta \tan \theta + \cos \theta = \sec \theta$ .
Hence proved $\sin \theta \tan \theta + \cos \theta = \sec \theta$ , identity verified.
So, the correct answer is “ $\sin \theta \tan \theta + \cos \theta = \sec \theta$ ”.

Note : Whenever you are asked to verify or prove any trigonometric equation or identity, you need to use all the properties that can be used here in order to simplify any one side of the equation and somehow bring it equal to the other side. You need to memorize the properties used this solution, such as inverse of cosine is equal to secant, ${\sin ^2}\theta + {\cos ^2}\theta = 1$, $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ . Use these properties appropriately to find the solution which is required.