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Question: Prove that trigonometric equation \(\dfrac{\tan \theta }{\sec \theta +1}=\dfrac{\sec \theta -1}{\tan...

Prove that trigonometric equation tan⁡θsec⁡θ+1=sec⁡θ−1tan⁡θ\dfrac{\tan \theta }{\sec \theta +1}=\dfrac{\sec \theta -1}{\tan \theta }.

Explanation

Solution

Hint: The given question is related to trigonometric identities. Try to recall the formulae related to the relationship between sine, cosine, tangent, and secant of an angle.

Complete step-by-step answer:

Before proceeding with the problem, first, let’s see the formulae used to solve the given problem.
tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta }
sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta }
1−cos⁡2θ=sin⁡2θ1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta
We need to prove that tan⁡θsec⁡θ+1=sec⁡θ−1tan⁡θ\dfrac{\tan \theta }{\sec \theta +1}=\dfrac{\sec \theta -1}{\tan \theta }.
First, we will consider the left-hand side of the equation. The left-hand side of the equation is given as tan⁡θsec⁡θ+1\dfrac{\tan \theta }{\sec \theta +1} . We know tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } and sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta } . So, the left-hand side of the equation becomes sin⁡θcos⁡θ1cos⁡θ+1\dfrac{\dfrac{\sin \theta }{\cos \theta }}{\dfrac{1}{\cos \theta }+1} .
⇒LHS=sin⁡θcos⁡θ1+cos⁡θcos⁡θ\Rightarrow LHS=\dfrac{\dfrac{\sin \theta }{\cos \theta }}{\dfrac{1+\cos \theta }{\cos \theta }}
⇒LHS=sin⁡θ1+cos⁡θ\Rightarrow LHS=\dfrac{\sin \theta }{1+\cos \theta }
Now, we know that the value of a fraction does not change on multiplying and dividing the fraction by the same number, except 00 .
So, LHS=sin⁡θ1+cos⁡θ×1−cos⁡θ1−cos⁡θLHS=\dfrac{\sin \theta }{1+\cos \theta }\times \dfrac{1-\cos \theta }{1-\cos \theta } .
⇒LHS=sin⁡θ(1−cos⁡θ)1−cos⁡2θ\Rightarrow LHS=\dfrac{\sin \theta \left( 1-\cos \theta \right)}{1-{{\cos }^{2}}\theta }
Now, we know that the value of 1−cos⁡2θ1-{{\cos }^{2}}\theta is equal to sin⁡2θ{{\sin }^{2}}\theta . So, the value of the left-hand side of the equation becomes sin⁡θ(1−cos⁡θ)sin⁡2θ\dfrac{\sin \theta \left( 1-\cos \theta \right)}{{{\sin }^{2}}\theta } .
⇒LHS=1−cos⁡θsin⁡θ\Rightarrow LHS=\dfrac{1-\cos \theta }{\sin \theta }
So, the value of the left-hand side of the equation is equal to 1−cos⁡θsin⁡θ\dfrac{1-\cos \theta }{\sin \theta } .
Now, we will consider the right-hand side of the equation. The right-hand side of the equation is given as sec⁡θ−1tan⁡θ\dfrac{\sec \theta -1}{\tan \theta } . We know tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } and sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta } . So, the right-hand side of the equation becomes 1cos⁡θ−1sin⁡θcos⁡θ\dfrac{\dfrac{1}{\cos \theta }-1}{\dfrac{\sin \theta }{\cos \theta }} .
⇒RHS=1−cos⁡θcos⁡θsin⁡θcos⁡θ\Rightarrow RHS=\dfrac{\dfrac{1-\cos \theta }{\cos \theta }}{\dfrac{\sin \theta }{\cos \theta }}
⇒RHS=1−cos⁡θsin⁡θ\Rightarrow RHS=\dfrac{1-\cos \theta }{\sin \theta }
So, the value of the right-hand side of the equation is equal to 1−cos⁡θsin⁡θ\dfrac{1-\cos \theta }{\sin \theta } . Clearly, the values of the left-hand side of the equation and the right-hand side of the equation are the same, i.e. LHS=RHS. Hence, proved.

Note: Students generally get confused and write 1+cos⁡2θ=sin⁡2θ1+{{\cos }^{2}}\theta ={{\sin }^{2}}\theta instead of 1−cos⁡2θ=sin⁡2θ1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta , which is wrong. These formulae should be properly remembered without any mistake as confusion in the formula can result in getting a wrong answer.