Question

Prove that \[\dfrac{1}{{\cos ec\theta + 1}} + \dfrac{1}{{\cos ec\theta - 1}} = 2\sec \theta \tan \theta \] ?

Answer 1

Hint: The given question deals with proving a trigonometric equality using the basic and simple trigonometric formulae and identities such as $\cos ecx = \dfrac{1}{{\sin x}}$. Basic algebraic rules and trigonometric identities are to be kept in mind while simplifying the given problem and proving the result given to us.

Complete answer:
For proving the desired result, we need to have a good grip over the basic trigonometric formulae and identities.
Now, we need to make the left and right sides of the equation equal.
L.H.S. $ = \dfrac{1}{{\cos ec\theta + 1}} + \dfrac{1}{{\cos ec\theta - 1}}$
So, we will convert all the trigonometric functions into sine and cosine using trigonometric formulae and identities. So, using the trigonometric formula $\cos ecx = \dfrac{1}{{\sin x}}$, we get,
$ = \dfrac{1}{{\dfrac{1}{{\sin \theta }} + 1}} + \dfrac{1}{{\dfrac{1}{{\sin \theta }} - 1}}$
Taking the LCM in the denominators, we get,
$ = \dfrac{1}{{\dfrac{{1 + \sin \theta }}{{\sin \theta }}}} + \dfrac{1}{{\dfrac{{1 - \sin \theta }}{{\sin \theta }}}}$
Simplifying the expression,
$ = \dfrac{{\sin \theta }}{{1 + \sin \theta }} + \dfrac{{\sin \theta }}{{1 - \sin \theta }}$
Multiplying the numerator and denominator of the first term by $\left( {1 - \sin \theta } \right)$ and that of the second term by $\left( {1 + \sin \theta } \right)$.
So, we get,
$ = \dfrac{{\sin \theta }}{{1 + \sin \theta }} \times \left( {\dfrac{{1 - \sin \theta }}{{1 - \sin \theta }}} \right) + \dfrac{{\sin \theta }}{{1 - \sin \theta }} \times \left( {\dfrac{{1 + \sin \theta }}{{1 + \sin \theta }}} \right)$
Using the algebraic identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$,
$ = \dfrac{{\sin \theta \left( {1 - \sin \theta } \right)}}{{1 - {{\sin }^2}\theta }} + \dfrac{{\sin \theta \left( {1 + \sin \theta } \right)}}{{1 - {{\sin }^2}\theta }}$
Simplifying the expression, we get,
\[ = \dfrac{{\sin \theta - {{\sin }^2}\theta }}{{1 - {{\sin }^2}\theta }} + \dfrac{{\sin \theta + {{\sin }^2}\theta }}{{1 - {{\sin }^2}\theta }}\]
Adding the numerator directly as the denominators of both the rational expression is same.
\[ = \dfrac{{\sin \theta + \sin \theta }}{{1 - {{\sin }^2}\theta }}\]
Now, applying the trigonometric identity \[{\sin ^2}x + {\cos ^2}x = 1\] in the common denominator, we get,
\[ = \dfrac{{2\sin \theta }}{{{{\cos }^2}\theta }}\]
Now, we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and $\sec x = \dfrac{1}{{\cos x}}$
\[ = 2\sec \theta \tan \theta \]
Now, R.H.S \[ = 2\sec \theta \tan \theta \]
As the left side of the equation is equal to the right side of the equation, we have,
\[\dfrac{1}{{\cos ec\theta + 1}} + \dfrac{1}{{\cos ec\theta - 1}} = 2\sec \theta \tan \theta \]
Hence, Proved.

Note:
Given problem deals with Trigonometric functions. For solving such problems, trigonometric formulae and identities such as $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and \[{\sin ^2}x + {\cos ^2}x = 1\] should be used. We also need knowledge of algebraic rules and identities to simplify the expression. Definitions of the trigonometric functions such as secant $\sec x = \dfrac{1}{{\cos x}}$, cosecant $\cos ecx = \dfrac{1}{{\sin x}}$ and tangent are essential for solving the problem.