evaluate the expression.\n\n\\( \\sin \\left( \\sin ^ { - 1 } \\left( - \\frac { 3 } { 5 } \\right) + \\tan…

evaluate the expression.\n\n\\( \\sin \\left( \\sin ^ { - 1 } \\left( - \\frac { 3 } { 5 } \\right) + \\tan ^ { - 1 } ( - 3 ) \\right) \\)\n\n\\( \\frac { 9 \\sqrt { 10 } } { 50 } \\)\n\\( \\frac { 3 \\sqrt { 10 } } { 10 } \\)\n\\( - \\frac { 9 \\sqrt { 10 } } { 50 } \\)\n\\( - \\frac { 3 \\sqrt { 10 } } { 10 } \\)

evaluate the expression.\n\n\\( \\sin \\left( \\sin ^ { - 1 } \\left( - \\frac { 3 } { 5 } \\right) + \\tan ^ { - 1 } ( - 3 ) \\right) \\)\n\n\\( \\frac { 9 \\sqrt { 10 } } { 50 } \\)\n\\( \\frac { 3 \\sqrt { 10 } } { 10 } \\)\n\\( - \\frac { 9 \\sqrt { 10 } } { 50 } \\)\n\\( - \\frac { 3 \\sqrt { 10 } } { 10 } \\)

Answer

Explanation:

Step1: Use the formula $\sin(A + B)=\sin A\cos B+\cos A\sin B$

Let $A=\sin^{-1}\left(-\frac{3}{5}\right)$ and $B = \tan^{-1}(-3)$. Then $\sin\left(\sin^{-1}\left(-\frac{3}{5}\right)+\tan^{-1}(-3)\right)=\sin\left(\sin^{-1}\left(-\frac{3}{5}\right)\right)\cos\left(\tan^{-1}(-3)\right)+\cos\left(\sin^{-1}\left(-\frac{3}{5}\right)\right)\sin\left(\tan^{-1}(-3)\right)$

Step2: Find the values of $\sin\left(\sin^{-1}\left(-\frac{3}{5}\right)\right)$, $\cos\left(\sin^{-1}\left(-\frac{3}{5}\right)\right)$, $\cos\left(\tan^{-1}(-3)\right)$ and $\sin\left(\tan^{-1}(-3)\right)$

  • By the property of inverse - sine function, if $y = \sin^{-1}x$, then $\sin y=x$ and $-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}$. So, $\sin\left(\sin^{-1}\left(-\frac{3}{5}\right)\right)=-\frac{3}{5}$.
  • Using the identity $\sin^{2}\theta+\cos^{2}\theta = 1$, if $\theta=\sin^{-1}\left(-\frac{3}{5}\right)$, then $\cos\theta=\sqrt{1-\sin^{2}\theta}$. Since $\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ and $\sin\theta=-\frac{3}{5}<0$, $\cos\theta=\sqrt{1-\left(-\frac{3}{5}\right)^{2}}=\frac{4}{5}$.
  • If $y = \tan^{-1}(-3)$, then $\tan y=-3=\frac{\sin y}{\cos y}$ and $-\frac{\pi}{2}<y<\frac{\pi}{2}$. Also, $\sin^{2}y+\cos^{2}y = 1$. Substituting $\sin y=-3\cos y$ into $\sin^{2}y+\cos^{2}y = 1$, we get $(-3\cos y)^{2}+\cos^{2}y = 1$, $9\cos^{2}y+\cos^{2}y = 1$, $10\cos^{2}y = 1$, $\cos y=\frac{1}{\sqrt{10}}$ (because when $\tan y=-3<0$ and $y\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, $\cos y>0$) and $\sin y=-\frac{3}{\sqrt{10}}$.

Step3: Substitute the values into the formula

[ \begin{align*} &\sin\left(\sin^{-1}\left(-\frac{3}{5}\right)\right)\cos\left(\tan^{-1}(-3)\right)+\cos\left(\sin^{-1}\left(-\frac{3}{5}\right)\right)\sin\left(\tan^{-1}(-3)\right)\ =&\left(-\frac{3}{5}\right)\times\frac{1}{\sqrt{10}}+\frac{4}{5}\times\left(-\frac{3}{\sqrt{10}}\right)\ =&-\frac{3 + 12}{5\sqrt{10}}\ =&-\frac{15}{5\sqrt{10}}\ =&-\frac{3}{\sqrt{10}}\ =&-\frac{3\sqrt{10}}{10} \end{align*} ]

Answer:

D. $-\frac{3\sqrt{10}}{10}$