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  • Need solution for RD Sharma maths class 12 Chapter Inverse Trigonometric functions exercise 3.8 question 1 sub question (ix)

Need solution for RD Sharma maths class 12 Chapter Inverse Trigonometric functions exercise 3.8 question 1 sub question (ix)

Answers (1)

\frac{7}{25}

Hint:

As we know that the range of \left(\cos ^{-1} x\right) \text { is } x \text { for } x \in[-1,1]

Given:

We have

                \cos \left(\tan ^{-1}\left(\frac{24}{7}\right)\right)

Solution:

In place of \cos ^{-1} x  here is \tan ^{-1} x

So, we convert \tan ^{-1} x  into \cos ^{-1} x .

Let’s suppose that,

\begin{aligned} &\tan ^{-1} \left (\frac{24}{7} \right )=\alpha \\ &\tan \alpha=\frac{24}{7} \end{aligned}

Let, in \Delta ABC,  angle CAB = \alpha and right angle at B.

\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(AC)^{2}=7^{2}+(24)^{2} \\ &(A C)^{2}=625 \end{aligned}

AC=\pm 25                [we will ignore the -ve sign because AB is a length and it can’t be -ve]

AC=25

\begin{aligned} &\cos \left(\tan ^{-1}\left(\frac{24}{7}\right)\right)=\cos \alpha \\ &\cos \alpha=\frac{7}{25} \\ &\cos \left(\tan ^{-1}\left(\frac{24}{7}\right)\right)=\frac{7}{25} \end{aligned}

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