Solveeit Logo

Question

Question: Prove that \[\dfrac{1}{{\cos ec\theta + 1}} + \dfrac{1}{{\cos ec\theta - 1}} = 2\sec \theta \tan \th...

Prove that 1cosecθ+1+1cosecθ1=2secθtanθ\dfrac{1}{{\cos ec\theta + 1}} + \dfrac{1}{{\cos ec\theta - 1}} = 2\sec \theta \tan \theta ?

Explanation

Solution

The given question deals with proving a trigonometric equality using the basic and simple trigonometric formulae and identities such as cosecx=1sinx\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. =1cosecθ+1+1cosecθ1 = \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 cosecx=1sinx\cos ecx = \dfrac{1}{{\sin x}}, we get,
=11sinθ+1+11sinθ1= \dfrac{1}{{\dfrac{1}{{\sin \theta }} + 1}} + \dfrac{1}{{\dfrac{1}{{\sin \theta }} - 1}}
Taking the LCM in the denominators, we get,
=11+sinθsinθ+11sinθsinθ= \dfrac{1}{{\dfrac{{1 + \sin \theta }}{{\sin \theta }}}} + \dfrac{1}{{\dfrac{{1 - \sin \theta }}{{\sin \theta }}}}
Simplifying the expression,
=sinθ1+sinθ+sinθ1sinθ= \dfrac{{\sin \theta }}{{1 + \sin \theta }} + \dfrac{{\sin \theta }}{{1 - \sin \theta }}
Multiplying the numerator and denominator of the first term by (1sinθ)\left( {1 - \sin \theta } \right) and that of the second term by (1+sinθ)\left( {1 + \sin \theta } \right).
So, we get,
=sinθ1+sinθ×(1sinθ1sinθ)+sinθ1sinθ×(1+sinθ1+sinθ)= \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 a2b2=(ab)(a+b){a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right),
=sinθ(1sinθ)1sin2θ+sinθ(1+sinθ)1sin2θ= \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,
=sinθsin2θ1sin2θ+sinθ+sin2θ1sin2θ= \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.
=sinθ+sinθ1sin2θ= \dfrac{{\sin \theta + \sin \theta }}{{1 - {{\sin }^2}\theta }}
Now, applying the trigonometric identity sin2x+cos2x=1{\sin ^2}x + {\cos ^2}x = 1 in the common denominator, we get,
=2sinθcos2θ= \dfrac{{2\sin \theta }}{{{{\cos }^2}\theta }}
Now, we know that tanx=sinxcosx\tan x = \dfrac{{\sin x}}{{\cos x}} and secx=1cosx\sec x = \dfrac{1}{{\cos x}}
=2secθtanθ= 2\sec \theta \tan \theta
Now, R.H.S =2secθtanθ= 2\sec \theta \tan \theta
As the left side of the equation is equal to the right side of the equation, we have,
1cosecθ+1+1cosecθ1=2secθtanθ\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 tanx=sinxcosx\tan x = \dfrac{{\sin x}}{{\cos x}} and sin2x+cos2x=1{\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 secx=1cosx\sec x = \dfrac{1}{{\cos x}}, cosecant cosecx=1sinx\cos ecx = \dfrac{1}{{\sin x}} and tangent are essential for solving the problem.