50\tan ( 3\tan ^ - 1( 1 \over 2 ) + 2\cos ^ - 1( 1 \over \sqrt 5 ) ) + 4\sqrt 2 \tan ( 1 \over 2\tan ^ - 1(2\sqrt 2 ) ) is equal to \underline\hspace2cm .

Inverse Trigonometric Functions PYQ — JEE Main 2022 · 29 Jun · Shift 1 · Q38 with Solution

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Correct answer: 29

Let $A=\tan^{-1}\left(\frac12\right),\qquad B=\cos^{-1}\left(\frac1{\sqrt5}\right).$ We need to evaluate $50\tan(3A+2B)+4\sqrt2\tan\left(\frac12\tan^{-1}(2\sqrt2)\right).$

First find $B$ .

Since $\cos B=\frac1{\sqrt5},$ we get $\sin B=\sqrt{1-\frac15}=\frac2{\sqrt5},$ so $\tan B=\frac{\sin B}{\cos B}=2.$ Thus, $B=\tan^{-1}(2)$ in the principal range.

Hence $2B=2\tan^{-1}(2).$

Now compute $\tan 3A$ where $\tan A=\frac12$ .

Using $\tan 3A=\frac{3t-t^3}{1-3t^2},\qquad t=\frac12,$ we get

=\frac{\frac32-\frac18}{1-\frac34} =\frac{\frac{11}{8}}{\frac14}=\frac{11}{2}.$$ So, $$\tan 3A=\frac{11}{2}.$$ --- 4. Compute $\tan 2B$ with $\tan B=2$. Using $$\tan 2B=\frac{2\tan B}{1-\tan^2 B},$$ we get $$\tan 2B=\frac{2\cdot 2}{1-4}=\frac4{-3}=-\frac43.$$ --- 5. Now evaluate $\tan(3A+2B)$. Using $$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y},$$ with $x=3A$, $y=2B$, $$\tan(3A+2B)=\frac{\frac{11}{2}-\frac43}{1-\frac{11}{2}\left(-\frac43\right)}.$$ Simplify numerator: $$\frac{11}{2}-\frac43=\frac{33-8}{6}=\frac{25}{6}.$$ Simplify denominator: $$1+\frac{44}{6}=1+\frac{22}{3}=\frac{25}{3}.$$ Therefore, $$\tan(3A+2B)=\frac{\frac{25}{6}}{\frac{25}{3}}=\frac12.$$ So the first term is $$50\tan(3A+2B)=50\cdot \frac12=25.$$ --- 6. Now evaluate $$4\sqrt2\tan\left(\frac12\tan^{-1}(2\sqrt2)\right).$$ Let $$\theta=\tan^{-1}(2\sqrt2),$$ so that $$\tan\theta=2\sqrt2.$$ We need $\tan\frac\theta2$. Use the identity $$\tan\frac\theta2=\frac{\sin\theta}{1+\cos\theta}.$$ From $\tan\theta=2\sqrt2$, take a right triangle with opposite $2\sqrt2$, adjacent $1$. Then hypotenuse is $$\sqrt{(2\sqrt2)^2+1^2}=\sqrt{8+1}=3.$$ So, $$\sin\theta=\frac{2\sqrt2}{3},\qquad \cos\theta=\frac13.$$ Hence, $$\tan\frac\theta2=\frac{\frac{2\sqrt2}{3}}{1+\frac13} =\frac{\frac{2\sqrt2}{3}}{\frac43} =\frac{\sqrt2}{2}. $$ Therefore the second term is $$4\sqrt2\cdot \frac{\sqrt2}{2}=4.$$ --- 7. Add both parts: $$25+4=29.$$ So the required integer is $$\boxed{29}. $$ --- 8. Comparison with stored answer: Stored correct answer = $29$. Our derived answer = $29$. So they agree.

More from Inverse Trigonometric Functions

View written solutionFree

Correct answer: 29

Let $A=\tan^{-1}\left(\frac12\right),\qquad B=\cos^{-1}\left(\frac1{\sqrt5}\right).$ We need to evaluate $50\tan(3A+2B)+4\sqrt2\tan\left(\frac12\tan^{-1}(2\sqrt2)\right).$

First find $B$ .

Since $\cos B=\frac1{\sqrt5},$ we get $\sin B=\sqrt{1-\frac15}=\frac2{\sqrt5},$ so $\tan B=\frac{\sin B}{\cos B}=2.$ Thus, $B=\tan^{-1}(2)$ in the principal range.

Hence $2B=2\tan^{-1}(2).$

Now compute $\tan 3A$ where $\tan A=\frac12$ .

Using $\tan 3A=\frac{3t-t^3}{1-3t^2},\qquad t=\frac12,$ we get

=\frac{\frac32-\frac18}{1-\frac34} =\frac{\frac{11}{8}}{\frac14}=\frac{11}{2}.$$ So, $$\tan 3A=\frac{11}{2}.$$ --- 4. Compute $\tan 2B$ with $\tan B=2$. Using $$\tan 2B=\frac{2\tan B}{1-\tan^2 B},$$ we get $$\tan 2B=\frac{2\cdot 2}{1-4}=\frac4{-3}=-\frac43.$$ --- 5. Now evaluate $\tan(3A+2B)$. Using $$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y},$$ with $x=3A$, $y=2B$, $$\tan(3A+2B)=\frac{\frac{11}{2}-\frac43}{1-\frac{11}{2}\left(-\frac43\right)}.$$ Simplify numerator: $$\frac{11}{2}-\frac43=\frac{33-8}{6}=\frac{25}{6}.$$ Simplify denominator: $$1+\frac{44}{6}=1+\frac{22}{3}=\frac{25}{3}.$$ Therefore, $$\tan(3A+2B)=\frac{\frac{25}{6}}{\frac{25}{3}}=\frac12.$$ So the first term is $$50\tan(3A+2B)=50\cdot \frac12=25.$$ --- 6. Now evaluate $$4\sqrt2\tan\left(\frac12\tan^{-1}(2\sqrt2)\right).$$ Let $$\theta=\tan^{-1}(2\sqrt2),$$ so that $$\tan\theta=2\sqrt2.$$ We need $\tan\frac\theta2$. Use the identity $$\tan\frac\theta2=\frac{\sin\theta}{1+\cos\theta}.$$ From $\tan\theta=2\sqrt2$, take a right triangle with opposite $2\sqrt2$, adjacent $1$. Then hypotenuse is $$\sqrt{(2\sqrt2)^2+1^2}=\sqrt{8+1}=3.$$ So, $$\sin\theta=\frac{2\sqrt2}{3},\qquad \cos\theta=\frac13.$$ Hence, $$\tan\frac\theta2=\frac{\frac{2\sqrt2}{3}}{1+\frac13} =\frac{\frac{2\sqrt2}{3}}{\frac43} =\frac{\sqrt2}{2}. $$ Therefore the second term is $$4\sqrt2\cdot \frac{\sqrt2}{2}=4.$$ --- 7. Add both parts: $$25+4=29.$$ So the required integer is $$\boxed{29}. $$ --- 8. Comparison with stored answer: Stored correct answer = $29$. Our derived answer = $29$. So they agree.

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