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Last updated date: 25th Nov 2023
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MVSAT Dec 2023

Evaluate the following
$\dfrac{{2\tan {{53}^0}}}{{\cot {{37}^0}}} - \dfrac{{\cot {{80}^0}}}{{\tan {{10}^0}}}$

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Hint- Try to bring the trigonometric functions in an unknown angle to some known ones by the use of identities.

First of all we will make numerator and denominator in the same terms, in order to cancel them out.
i.e. we will convert $\tan \theta $ in terms of $\cot \theta $ and vice-versa.
We know that
\[{\text{cot}}\left( {{{90}^0} - \theta } \right) = \tan \theta {\text{ & tan}}\left( {{{90}^0} - \theta } \right) = \cot \theta \]
Now according to the question
   \Rightarrow \dfrac{{2\tan {{53}^0}}}{{\cot \left( {{{90}^0} - {{37}^0}} \right)}} - \dfrac{{\cot {{80}^0}}}{{\tan \left( {{{90}^0} - {{10}^0}} \right)}} \\
   \Rightarrow \dfrac{{2\tan {{53}^0}}}{{\tan {{53}^0}}} - \dfrac{{\cot {{80}^0}}}{{\cot {{80}^0}}} \\
Canceling out common terms in numerator and denominator
   \Rightarrow 2\left( 1 \right) - \left( 1 \right) \\
   \Rightarrow 1 \\
Hence, value of the above term is equal to $1$

Note- Whenever there is an integer angle inside the trigonometric functions whose value is not known to us, always try to convert it using \[{90^0} - \theta \] so as to make terms easy to calculate.