Question

If \[\tan \theta +\cot \theta =5\], find the value of \[{{\tan }^{2}}\theta +{{\cot }^{2}}\theta \]

Answer 1

Hint: First of all take \[\tan \theta =x\]. So \[\cot \theta \] would be \[\dfrac{1}{x}\]. Now square both sides of the given equation and use \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] to get the desired value.

Complete Step-by-step answer:
We are given that \[\tan \theta +\cot \theta =5\]. Here, we have to find the value of the expression \[{{\tan }^{2}}\theta +{{\cot }^{2}}\theta \]. Let us consider the equation given in the question,
\[\tan \theta +\cot \theta =5....\left( i \right)\]
Let us assume \[\tan \theta =x\]. We know that \[\tan \theta \] and \[\cot \theta \] are reciprocal of each other that is \[\tan \theta =\dfrac{1}{\cot \theta }\] or \[\cot \theta =\dfrac{1}{\tan \theta }\]. Also, we have assumed that \[\tan \theta =x\]. So, we get,
\[\cot \theta =\dfrac{1}{\tan \theta }=\dfrac{1}{x}\]
So, now by substituting the value of \[\tan \theta =x\] and \[\cot \theta =\dfrac{1}{x}\] in equation (i), we get,
\[x+\dfrac{1}{x}=5\]
By squaring both the sides of the above equation, we get,
\[{{\left( x+\dfrac{1}{x} \right)}^{2}}={{\left( 5 \right)}^{2}}\]
\[{{\left( x+\dfrac{1}{x} \right)}^{2}}=25\]
We know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]. So by using this in the above equation and considering a = x and \[b=\dfrac{1}{x}\], we get,
\[{{\left( x \right)}^{2}}+{{\left( \dfrac{1}{x} \right)}^{2}}+2\left( x \right).\left( \dfrac{1}{x} \right)=25\]
By simplifying the above equation, and canceling the like terms, we get,
\[{{x}^{2}}+\dfrac{1}{{{x}^{2}}}+2=25\]
By subtracting 2 from both the sides of the above equation, we get,
\[{{x}^{2}}+\dfrac{1}{{{x}^{2}}}+2-2=25-2\]
\[{{x}^{2}}+\dfrac{1}{{{x}^{2}}}=23\]
Now by substituting \[x=\tan \theta \](as we have initially assumed) in the above equation, we get that
\[{{\tan }^{2}}\theta +\dfrac{1}{{{\tan }^{2}}\theta }=23\]
\[{{\tan }^{2}}\theta +{{\left( \dfrac{1}{\tan \theta } \right)}^{2}}=23\]
We know that \[\dfrac{1}{\tan \theta }=\cot \theta \]. By using this in the above equation, we get,
\[{{\tan }^{2}}\theta +{{\cot }^{2}}\theta =23\]
Hence, we have found that \[{{\tan }^{2}}\theta +{{\cot }^{2}}\theta =23\]

Note: In this question, some students find \[\tan \theta \] and \[\cot \theta \] individually. Though this is also correct, this method can get lengthy and lead to mistakes even if there is a small calculation mistake. So, it is better to use the above method if given trigonometric ratios are reciprocal to each other.