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Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise Multiple Chioce Questions Question 2 Maths Textbook Solution.

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Answer: \frac{3}{29}

Hint: Take tan function to RHS, so that the variables get free.

Given: \tan \left[\cos ^{-1} \frac{1}{5 \sqrt{2}}-\sin ^{-1} \frac{4}{\sqrt{17}}\right]

Solution: Let,\cos ^{-1} \frac{1}{5 \sqrt{2}}=y, \sin ^{-1} \frac{4}{\sqrt{17}}=z

            \begin{aligned} &\therefore \cos y=\frac{1}{5 \sqrt{2}} \Rightarrow \sin y=\frac{7}{5 \sqrt{2}} \Rightarrow \tan y=7 \\ &\quad \sin z=\frac{4}{\sqrt{17}} \Rightarrow \cos z=\frac{1}{\sqrt{17}} \Rightarrow \tan z=4 \\ &\therefore \tan \left(\cos ^{-1} \frac{1}{5 \sqrt{2}}-\sin ^{-1} \frac{4}{\sqrt{17}}\right)=\tan (y-z) \\ &=\frac{\tan y-\tan z}{1+\tan y \tan z}=\frac{7-4}{1+7 \times 4} \\ &=\frac{3}{29} \end{aligned}

Mention step how siny has been taken as 7/5undrt2  

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