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Question: Prove that \({{\left( \tan \theta +\dfrac{1}{\cos \theta } \right)}^{2}}+{{\left( \tan \theta -\dfra...

Prove that (tan⁡θ+1cos⁡θ)2+(tan⁡θ−1cos⁡θ)2=2(1+sin⁡2θ1−sin⁡2θ){{\left( \tan \theta +\dfrac{1}{\cos \theta } \right)}^{2}}+{{\left( \tan \theta -\dfrac{1}{\cos \theta } \right)}^{2}}=2\left( \dfrac{1+{{\sin }^{2}}\theta }{1-{{\sin }^{2}}\theta } \right).

Explanation

Solution

Hint: Expand the L.H.S term and simplify it and Use Trigonometric ratios and identities to get the required answer.

“Complete step-by-step answer:”
To prove, (tan⁡θ+1cos⁡θ)2+(tan⁡θ−1cos⁡θ)2=2(1+sin⁡2θ1−sin⁡2θ){{\left( \tan \theta +\dfrac{1}{\cos \theta } \right)}^{2}}+{{\left( \tan \theta -\dfrac{1}{\cos \theta } \right)}^{2}}=2\left( \dfrac{1+{{\sin }^{2}}\theta }{1-{{\sin }^{2}}\theta } \right)
Left hand side (L.H.S) =(tan⁡θ+1cos⁡θ)2+(tan⁡θ−1cos⁡θ)2={{\left( \tan \theta +\dfrac{1}{\cos \theta } \right)}^{2}}+{{\left( \tan \theta -\dfrac{1}{\cos \theta } \right)}^{2}}
We open the brackets by the formula (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab.
∴LHS=(tan⁡2θ+1cos⁡2θ+2tan⁡θcos⁡θ)+(tan⁡2θ+1cos⁡2θ−2tan⁡θcos⁡θ) ⇒LHS=2tan⁡2θ+2cos⁡2θ ⇒LHS=2(tan⁡2θ+1cos⁡2θ) \begin{aligned} & \therefore LHS=\left( {{\tan }^{2}}\theta +\dfrac{1}{{{\cos }^{2}}\theta }+2\tan \theta \cos \theta \right)+\left( {{\tan }^{2}}\theta +\dfrac{1}{{{\cos }^{2}}\theta }-2\tan \theta \cos \theta \right) \\\ & \Rightarrow LHS=2{{\tan }^{2}}\theta +\dfrac{2}{{{\cos }^{2}}\theta } \\\ & \Rightarrow LHS=2\left( {{\tan }^{2}}\theta +\dfrac{1}{{{\cos }^{2}}\theta } \right) \\\ \end{aligned}
Let us write tan⁡θ as sin⁡θcos⁡θ\tan \theta \ as\ \dfrac{\sin \theta }{\cos \theta }.
⇒LHS=2[(sin⁡θcos⁡θ)2+1cos⁡2θ] ⇒LHS=2(sin⁡2θcos⁡2θ+1cos⁡2θ) ⇒LHS=2(1+sin⁡2θcos⁡2θ) \begin{aligned} & \Rightarrow LHS=2\left[ {{\left( \dfrac{\sin \theta }{\cos \theta } \right)}^{2}}+\dfrac{1}{{{\cos }^{2}}\theta } \right] \\\ & \Rightarrow LHS=2\left( \dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }+\dfrac{1}{{{\cos }^{2}}\theta } \right) \\\ & \Rightarrow LHS=2\left( \dfrac{1+{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta } \right) \\\ \end{aligned}
We know that 1−sin⁡2θ=cos⁡2θ1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta .
∴LHS=2(1+sin⁡2θ1−sin⁡2θ) L.H.S=R.H.S \begin{aligned} & \therefore LHS=2\left( \dfrac{1+{{\sin }^{2}}\theta }{1-{{\sin }^{2}}\theta } \right) \\\ & L.H.S=R.H.S \\\ \end{aligned}
Hence proved.

Note: In questions like these, our aim is to correct the longer side, in this case L.H.S identical to the other side. Therefore, after every step, we try to manipulate the expression.
Formulae used:
(a+b)2=a2+2ab+b2 cos⁡2θ+sin⁡2θ=1 tan⁡2θ=sin⁡2θcos⁡2θ \begin{aligned} & {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\\ & {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \\\ & {{\tan }^{2}}\theta =\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta } \\\ \end{aligned}