Question

If \[\tan \theta + \sec \theta = x\], then prove that \[\sin \theta = \dfrac{{{x^2} - 1}}{{{x^2} + 1}}\]?

Answer 1

Hint: To solve this question first we first we multiply and divide by \[\sec \theta - \tan \theta \] on the left-hand side. Then we use trigonometry to eliminate some parts. And try to make an equation in \[\sec \theta - \tan \theta \]. Then from these equations try to find the value of \[\sec \theta \] and \[\tan \theta \]. Then try to express \[\sin \theta \] in terms of \[\sec \theta \] and \[\tan \theta \]. And put the values in the final equation and simplify that in order to get the value of \[\sin \theta \] and that is the final answer.

Complete step-by-step answer:
Given,
\[\tan \theta + \sec \theta = x\] ……(i)

On reciprocating the given equation.
\[\dfrac{1}{{\tan \theta + \sec \theta }} = \dfrac{1}{x}\]
On multiplying and divide by \[\sec \theta - \tan \theta \] in left hand side.
\[\dfrac{{\sec \theta - \tan \theta }}{{\left( {\tan \theta + \sec \theta } \right)\left( {\sec \theta - \tan \theta } \right)}} = \dfrac{1}{x}\]
On further calculating
\[\dfrac{{\sec \theta - \tan \theta }}{{{{\left( {\sec \theta } \right)}^2} - {{\left( {\tan \theta } \right)}^2}}} = \dfrac{1}{x}\]
Now on using the identity of trigonometry \[{\left( {\sec \theta } \right)^2} - {\left( {\tan \theta } \right)^2} = 1\]
\[\dfrac{{\sec \theta - \tan \theta }}{1} = \dfrac{1}{x}\]
\[\sec \theta - \tan \theta = \dfrac{1}{x}\] ……(ii)
Adding equation (i) and (ii) we get
\[2\sec \theta = x + \dfrac{1}{x}\]
Subtracting (ii) from (i) we get
\[2\tan \theta = x - \dfrac{1}{x}\]
Dividing last two equations
\[\dfrac{{tan\theta }}{{sec\theta }} = \dfrac{{x - \dfrac{1}{x}}}{{x + \dfrac{1}{x}}}\]
On taking LCM on denominator-
\[\dfrac{{tan\theta }}{{sec\theta }} = \dfrac{{\dfrac{{{x^2} - 1}}{x}}}{{\dfrac{{{x^2} + 1}}{x}}}\]
On solving the left hand side and simplifying the right hand side.
\[\sin \theta = \dfrac{{{x^2} - 1}}{{{x^2} + 1}}\]
Hence proved.

Note: We can solve this method by another method. Also in that method we start from the right-hand side and try to prove that equal to the left-hand side. We put the value of the x on the right-hand side and then simplify that and try to get \[\sin \theta \] in terms of x. That is our final answer. But this method is the long method and there are more chances to get the wrong answer and we must commit mistakes. In this method, students are not able to figure out how to make another equation and find the value of \[\sec \theta \] and \[\tan \theta \].