# Ex.8.3 Q4 Introduction to Trigonometry Solution - NCERT Maths Class 10

Go back to  'Ex.8.3'

## Question

If $$\tan A = \cot B\,,$$ prove that $$A + B = {90^0}.$$

## Text Solution

#### Reasoning:

$$\tan ({90^0} - \theta ) = \cot \theta$$

#### Steps:

Given that:

\begin{align}\,\tan A = \cot B & & (i)\end{align}

We know that,

$$\tan A = \cot \left( {{{90}^0} - A} \right)$$

By substituting this in equation (i) we get:

\begin{align} \cot \,({90^0} - A) &= \cot B\\ {90^0} - A &= B\\ A + B &= {90^0} \end{align}

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