Question

Find the value of \[\cot \left( {{90}^{\circ }}-\theta \right)\] from the given figure.
A. \[\sin \theta \]
B. \[\cos \theta \]
C. \[\tan \theta \]
D. None of these

Answer 1

Hint: Find the relations of trigonometric ratios \[\left( {{90}^{\circ }}-\theta \right)\] with the help of quadrants. Assume \[\Delta ABC\] lies in the first quadrant and find \[\cot \left( {{90}^{\circ }}-\theta \right)\] with the help of basic identities of trigonometry.

Complete step-by-step answer:

We can find relations among all the trigonometric ratios of \[\left( {{90}^{\circ }}-\theta \right)\].

The easy way to find the value can be done with the help of quadrants.
In the first quadrant, all values are positive.

In the second quadrant, \[\sin \theta \] and \[\cos ec\theta \] are positive.

In the third quadrant, \[tan\theta \] and \[\cot \theta \] are positive.

In the fourth quadrant, \[\cos \theta \] and \[sec\theta \] are positive.

From this we can make it clear that \[\left( {{90}^{\circ }}-\theta \right)\] lies in the first quadrant. So to evaluate \[\cot \left( {{90}^{\circ }}-\theta \right)\], let us assume that, \[\left( {{90}^{\circ }}-\theta \right)\] falls in the first quadrant where all values are positive.

This means that \[\Delta ABC\] is in the first quadrant.

When we have \[{{90}^{\circ }}\], ‘cot’ will become ‘tan’. These are basic identities of trigonometry.

And so in the first quadrant, the sign of ‘cot’ is positive.

Hence \[\cot \left( {{90}^{\circ }}-\theta \right)=\tan \theta \].

Option C is the correct answer.

Note: We know that by basic trigonometry,

\[\begin{align}

  & \cot \theta =\dfrac{1}{\tan \theta } \\

 & \therefore \cot ({{90}^{\circ }}-\theta )=\dfrac{1}{\tan ({{90}^{\circ }}-\theta )} \\

\end{align}\]

We know that \[tan({{90}^{\circ }}-\theta )=\cot \theta \]

\[\begin{align}

  & \therefore \cot ({{90}^{\circ }}-\theta )=\dfrac{1}{\cot \theta } \\
 & \therefore \cot (90-\theta )=\tan \theta \\

\end{align}\]

We know \[\begin{align}

  & \cot \theta =\dfrac{1}{\tan \theta }. \\

 & \therefore \tan \theta =\dfrac{1}{\cot \theta } \\
\end{align}\]