AP DSC School Assistant Math Model Test 3

Section: Subject Specific

Question : 102 of 160

Marks: +1, -0

\rm \sin^{-1}\frac{4}{5}-\sin^{-1}\frac{5}{13}=?

Solution:

Concept:

Use the standard identity for the difference of inverse sines:

\sin^{-1}x - \sin^{-1}y = \sin^{-1}\!\big[x\sqrt{1-y^2} - y\sqrt{1-x^2}\,\big]

, where

x,y \in [0,1]

and the expression lies in the principal range.

Explanation:

Apply the formula with

x = \frac{4}{5}

and

y = \frac{5}{13}

\sin^{-1}\frac{4}{5} - \sin^{-1}\frac{5}{13} = \sin^{-1}\!\left[\frac{4}{5}\sqrt{1-\left(\frac{5}{13}\right)^2} - \frac{5}{13}\sqrt{1-\left(\frac{4}{5}\right)^2}\right]

Compute each square root:

\sqrt{1-\left(\frac{5}{13}\right)^2} = \sqrt{1-\frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}

\sqrt{1-\left(\frac{4}{5}\right)^2} = \sqrt{1-\frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}

Substitute these values:

\frac{4}{5}\cdot\frac{12}{13} - \frac{5}{13}\cdot\frac{3}{5} = \frac{48}{65} - \frac{15}{65} = \frac{33}{65}

Thus the expression equals

\sin^{-1}\frac{33}{65}

Answer:

\sin^{-1}\frac{33}{65}

(Option C).

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