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Question: If \[\tan \dfrac{A}{2} =r\] then the value of \[\left( \sec A+\tan A\right) \] is equal to A) \[\d...

If tan⁡A2=r\tan \dfrac{A}{2} =r then the value of (sec⁡A+tan⁡A)\left( \sec A+\tan A\right) is equal to
A) 2+r2−r\dfrac{2+r}{2-r}
B) 2−r2+r\dfrac{2-r}{2+r}
C) 1+r1−r\dfrac{1+r}{1-r}
D) 1−r1+r\dfrac{1-r}{1+r}

Explanation

Solution

Hint: In this question it is given that we have to find the value of (sec⁡A+tan⁡A)\left( \sec A+\tan A\right) , where tan⁡A2=r\tan \dfrac{A}{2} =r. So to find the solution we have to write sec⁡A\sec A, tan⁡A \tan A as 1cos⁡A\dfrac{1}{\cos A}, sin⁡Acos⁡A\dfrac{\sin A}{\cos A} respectively. Also we have to change the angle AA to A2\dfrac{A}{2}, because as we know that tan⁡A2=r\tan \dfrac{A}{2} =r and then after simplification we will get our required solution.
Complete step-by-step solution:
Let the given expression as
S=(sec⁡A+tan⁡A)S=\left( \sec A+\tan A\right)
=(1cos⁡A+sin⁡Acos⁡A)=\left( \dfrac{1}{\cos A} +\dfrac{\sin A}{\cos A} \right)
=(1+sin⁡Acos⁡A)=\left( \dfrac{1+\sin A}{\cos A} \right)
Now as we know that,
cos⁡θ=cos⁡2θ2−sin⁡2θ2\cos \theta =\cos^{2} \dfrac{\theta }{2} -\sin^{2} \dfrac{\theta }{2}, and
sin⁡θ=2sin⁡θ2cos⁡θ2\sin \theta =2\sin \dfrac{\theta }{2} \cos \dfrac{\theta }{2}
so by using this formula, we get,
S=1+2sin⁡A2cos⁡A2cos⁡2A2−sin⁡2A2S=\dfrac{1+2\sin \dfrac{A}{2} \cos \dfrac{A}{2} }{\cos^{2} \dfrac{A}{2} -\sin^{2} \dfrac{A}{2} }
=sin⁡2A2+cos⁡2A2+2sin⁡A2cos⁡A2cos⁡2A2−sin⁡2A2=\dfrac{\sin^{2} \dfrac{A}{2} +\cos^{2} \dfrac{A}{2} +2\sin \dfrac{A}{2} \cos \dfrac{A}{2} }{\cos^{2} \dfrac{A}{2} -\sin^{2} \dfrac{A}{2} } [ since, sin⁡2A2+cos⁡2A2=1\sin^{2} \dfrac{A}{2} +\cos^{2} \dfrac{A}{2} =1]
As we know that,
a2+2ab+b2=(a+b)2a^{2}+2ab+b^{2}=\left( a+b\right)^{2} .......(1) and
a2−b2=(a+b)(a−b)a^{2}-b^{2}=\left( a+b\right) \left( a-b\right) .....(2),
so by applying identity (1) in numerator and identity (2) in denominator we get,
S=(cos⁡A2+sin⁡A2)2(cos⁡A2+sin⁡A2)(cos⁡A2−sin⁡A2)S=\dfrac{\left( \cos \dfrac{A}{2} +\sin \dfrac{A}{2} \right)^{2} }{\left( \cos \dfrac{A}{2} +\sin \dfrac{A}{2} \right) \left( \cos \dfrac{A}{2} -\sin \dfrac{A}{2} \right) } [ taking a=cos⁡A2a=\cos \dfrac{A}{2} & b=sin⁡A2b=\sin \dfrac{A}{2}]
=(cos⁡A2+sin⁡A2)(cos⁡A2−sin⁡A2)=\dfrac{\left( \cos \dfrac{A}{2} +\sin \dfrac{A}{2} \right) }{\left( \cos \dfrac{A}{2} -\sin \dfrac{A}{2} \right) }
Now dividing numerator and denominator by cos⁡A2\cos \dfrac{A}{2}, we get,
S=1+(sin⁡A2cos⁡A2)1−(sin⁡A2cos⁡A2)S=\dfrac{1+\left( \dfrac{\sin \dfrac{A}{2} }{\cos \dfrac{A}{2} } \right) }{1-\left( \dfrac{\sin \dfrac{A}{2} }{\cos \dfrac{A}{2} } \right) }
=1+tan⁡A21−tan⁡A2=\dfrac{1+\tan \dfrac{A}{2} }{1-\tan \dfrac{A}{2} }
=1+r1−r=\dfrac{1+r}{1-r} [since, tan⁡A2=r\tan \dfrac{A}{2} =r]
Hence, the correct option is option C.
Note: While simplifying a big expression, try to express it in terms of one or two basic trigonometric functions, like we have transformed the above expression in terms of cosinecosine and sinesine, also try to find an order in the problem to apply trigonometric identities, properties and transformations.