Question

if sin∅=16÷65 then tan ∅=​

Answer 1

Answer:

To prove:

\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)+\sin ^{-1}\left(\frac{16}{65}\right)=\frac{\pi}{2}

Solution:

\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)+\sin ^{-1}\left(\frac{16}{65}\right)=\frac{\pi}{2}

\Rightarrow \sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{16}{65}\right)

\Rightarrow \sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)=\cos ^{-1}\left(\frac{16}{65}\right)

We just prove \bold{\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)=\cos ^{-1}\left(\frac{16}{65}\right)}

We take left hand side =\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)

Let \sin ^{-1}\left(\frac{4}{5}\right)=x \text { and } \sin ^{-1}\left(\frac{5}{13}\right)=y

\Rightarrow \sin x=\frac{4}{5} \text { and } \sin y=\frac{5}{13}

Now, \cos x=\sqrt{1-\sin ^{2} x}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}

Now, \cos y=\sqrt{1-\sin ^{2} y}=\sqrt{1-\frac{25}{169}}=\frac{12}{13}

Now, \bold{\cos (x+y)=\cos x \cdot \cos y-\sin x \cdot \sin y}

\Rightarrow \cos (x+y)=\frac{3}{5} \times \frac{12}{13}-\frac{4}{5} \times \frac{5}{13}

\Rightarrow \cos (x+y)=\frac{36}{65}-\frac{20}{65}

\Rightarrow \cos (x+y)=\frac{16}{65}

\Rightarrow x+y=\cos ^{-1}\left(\frac{16}{65}\right)

\bold{\Rightarrow \sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)=\cos ^{-1}\left(\frac{16}{65}\right)} = Right hand side.

Hence proved.