3.13 Property 13

Question 4. Solve: $\cos \left[ {{{\sin }^{ - 1}}\frac{3}{5} + {{\sin }^{ - 1}}\frac{5}{{13}}} \right]$

Solution: Let $$\begin{equation} \begin{aligned} {\sin ^{ - 1}}\frac{3}{5} = \theta \\ \sin \theta = \frac{3}{5} \\\end{aligned} \end{equation} $$

Let $$\begin{equation} \begin{aligned} {\sin ^{ - 1}}\frac{5}{{13}} = \phi \\ \sin \phi = \frac{5}{{13}} \\\end{aligned} \end{equation} $$

The given expression becomes $\cos [\theta + \phi ]$ so using $$\cos [\theta + \phi ] = \cos \theta \cos \phi - \sin \theta \sin \phi $$

$$\begin{equation} \begin{aligned} \frac{4}{5} \times \frac{{12}}{{13}} - \frac{3}{5} \times \frac{5}{{13}} \\ \frac{{48}}{{65}} - \frac{{15}}{{65}} \\ \frac{{33}}{{65}} \\\end{aligned} \end{equation} $$

Question 5. Prove that ${{{\tan }^{ - 1}}\frac{1}{7} + {{\tan }^{ - 1}}\frac{1}{{13}} = {{\tan }^{ - 1}}\frac{2}{9}}$

Solution: Applying the formula, ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right)$

$${\tan ^{ - 1}}\frac{1}{7} + {\tan ^{ - 1}}\frac{1}{{13}} = {\tan ^{ - 1}}\left( {\frac{{\frac{1}{7} + \frac{1}{{13}}}}{{1 - \frac{1}{7} \times \frac{1}{{13}}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{20}}{{90}}} \right) = {\tan ^{ - 1}}\frac{2}{9}$$

Question 6.Prove that: ${\tan ^{ - 1}}\sqrt x = \frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right)$

Solution: Let $\sqrt x = \tan \theta $

then $$LHS = \theta $$

RHS: $$\frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$$

We know that $\cos 2\theta = \left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$$\frac{1}{2}{\cos ^{ - 1}}(\cos 2\theta )$$

Using Property ${\cos ^{ - 1}}(\cos x) = x$, we have

$$\frac{1}{2} \times 2\theta $$

$$ = \theta $$

Question 7. Solve: $${\tan ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right) = \frac{1}{2}{\tan ^{ - 1}}x,x > 0$$

Solution: $$\begin{equation} \begin{aligned} x = \tan \theta \\ {\tan ^{ - 1}}\left( {\frac{{1 - \tan \theta }}{{1 + \tan \theta }}} \right) = \frac{1}{2}{\tan ^{ - 1}}\left( {\tan \theta } \right) \\\end{aligned} \end{equation} $$

We know that $\tan \frac{\pi }{4} = 1$, so putting this in place of $1$ we get $${\tan ^{ - 1}}\left( {\frac{{\tan \frac{\pi }{4} - \tan \theta }}{{1 + \tan \frac{\pi }{4}\tan \theta }}} \right) = \frac{1}{2}{\tan ^{ - 1}}\left( {\tan \theta } \right)$$

Using property ${\tan ^{ - 1}}\left( {\tan x} \right) = x$

$$\begin{equation} \begin{aligned} {\tan ^{ - 1}}\left( {\frac{{\tan \frac{\pi }{4} - \tan \theta }}{{1 + \tan \frac{\pi }{4}\tan \theta }}} \right) = \frac{1}{2}{\tan ^{ - 1}}\left( {\tan \theta } \right) \\ {\tan ^{ - 1}}\left[ {\tan \left\{ {\frac{\pi }{4} - \theta } \right\}} \right] = \frac{1}{2}\theta \\ \frac{\pi }{4} - \theta = \frac{1}{2}\theta \\ \frac{3}{2}\theta = \frac{\pi }{4} \\ \theta = \frac{\pi }{6} \\ x = \tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }} \\\end{aligned} \end{equation} $$

Question 8. Show that $${\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right) = \frac{\pi }{4} + \frac{1}{2}{\cos ^{ - 1}}\left( {{x^2}} \right)$$

Solution: Let $$\begin{equation} \begin{aligned} {x^2} = \cos 2\theta \\ 2\theta = {\cos ^{ - 1}}{x^2} \\ \theta = \frac{1}{2}{\cos ^{ - 1}}{x^2} \\\end{aligned} \end{equation} $$

Put ${x^2} = \cos 2\theta $ in LHS of equation, we have

$$\begin{equation} \begin{aligned} {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right) \\ {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + \cos 2\theta } + \sqrt {1 - \cos 2\theta } }}{{\sqrt {1 + \cos 2\theta } - \sqrt {1 - \cos 2\theta } }}} \right) \\ {\tan ^{ - 1}}\left( {\frac{{\sqrt {2{{\cos }^2}\theta } + \sqrt {2{{\sin }^2}\theta } }}{{\sqrt {2{{\cos }^2}\theta } - \sqrt {2{{\sin }^2}\theta } }}} \right) \\ {\tan ^{ - 1}}\left( {\frac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }}} \right) \\\end{aligned} \end{equation} $$

Dividing numerator and denominator by ${\cos \theta }$,we get $$\begin{equation} \begin{aligned} {\tan ^{ - 1}}\left( {\frac{{1 + \tan \theta }}{{1 - \tan \theta }}} \right) \\ {\tan ^{ - 1}}\left[ {\tan \left( {\frac{\pi }{4} + \theta } \right)} \right] \\ \left( {\frac{\pi }{4} + \theta } \right) \\ = \frac{\pi }{4} + \frac{1}{2}{\cos ^{ - 1}}\left( {{x^2}} \right) \\\end{aligned} \end{equation} $$

Question 9. If $x$, $y$ and $z$ are in A.P and ${\tan ^{ - 1}}x$, ${\tan ^{ - 1}}y$ and ${\tan ^{ - 1}}z$ are also in AP, then show that $x=y=z$

Solution: Since $x$, $y$ and $z$ are in A.P so $2y=x+z$

Also ${\tan ^{ - 1}}x$, ${\tan ^{ - 1}}y$ and ${\tan ^{ - 1}}z$ are in AP, so

$$\begin{equation} \begin{aligned} 2{\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}z \\ {\tan ^{ - 1}}\left( {\frac{{2y}}{{1 - {y^2}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{x + z}}{{1 - xz}}} \right) \\ \frac{{2y}}{{1 - {y^2}}} = \frac{{x + z}}{{1 - xz}} \\ (x + z)(1 - {y^2}) = (1 - xz)(2y) \\\end{aligned} \end{equation} $$

Put $2y=x+z$, we get

$$\begin{equation} \begin{aligned} (1 - {y^2}) = 1 - xz \\ {y^2} = xz \\\end{aligned} \end{equation} $$

So $x$, $y$ and $z$ are in A.P as well as in GP so $x=y=z$

Question 10. Find the value of $\cot \left\{ {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{k = 1}^n {2k} } \right)} } \right\}$

Solution: $$\begin{equation} \begin{aligned} \cot \left\{ {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{k = 1}^n {2k} } \right)} } \right\} \\ \cot \left\{ {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + 2 + 4 + 6 + ... + n} \right)} } \right\} \\\end{aligned} \end{equation} $$

$$\cot \left\{ {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}(1 + n(n + 1))} } \right\}$$

$$\cot \left\{ {\sum\limits_{n = 1}^{23} {{{\tan }^{ - 1}}\frac{1}{{1 + n(n + 1)}}} } \right\}$$

$$\cot \left\{ {\sum\limits_{n = 1}^{23} {{{\tan }^{ - 1}}\frac{{(n + 1) - n}}{{1 + n(n + 1)}}} } \right\}$$

$$\cot \left\{ {\sum\limits_{n = 1}^{23} {{{\tan }^{ - 1}}(n + 1) - {{\tan }^{ - 1}}n} } \right\}$$

$$\begin{equation} \begin{aligned} \cot \left\{ {({{\tan }^{ - 1}}2 - {{\tan }^{ - 1}}1) + ({{\tan }^{ - 1}}3 - {{\tan }^{ - 1}}2) + ({{\tan }^{ - 1}}4 - {{\tan }^{ - 1}}3)...} \right\} \\ \cot \left\{ {{{\tan }^{ - 1}}24 - {{\tan }^{ - 1}}1} \right\} \\ \cot \left\{ {{{\tan }^{ - 1}}\frac{{24 - 1}}{{1 + 24}}} \right\} \\ \cot \left\{ {{{\tan }^{ - 1}}\frac{{23}}{{25}}} \right\} \\ \cot \left\{ {{{\cot }^{ - 1}}\frac{{25}}{{23}}} \right\} \\ \frac{{25}}{{23}} \\\end{aligned} \end{equation} $$

Question 11. The value of $x$ for which $$\sin \left[ {{{\cot }^{ - 1}}(1 + x)} \right] = \cos ({\tan ^{ - 1}}x)$$ is

Solution: We know that $$\begin{equation} \begin{aligned} {\cot ^{ - 1}}x = {\sin ^{ - 1}}\left( {\frac{1}{{\sqrt {1 + {x^2}} }}} \right) \\ {\tan ^{ - 1}}x = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt {1 + {x^2}} }}} \right) \\\end{aligned} \end{equation} $$

So $$\begin{equation} \begin{aligned} \sin \left[ {{{\sin }^{ - 1}}\left( {\frac{1}{{\sqrt {1 + {{(1 + x)}^2}} }}} \right)} \right] = \cos \left[ {{{\cos }^{ - 1}}\left( {\frac{1}{{\sqrt {1 + {x^2}} }}} \right)} \right] \\ \frac{1}{{\sqrt {1 + {{(1 + x)}^2}} }} = \frac{1}{{\sqrt {1 + {x^2}} }} \\ 1 + {(1 + x)^2} = 1 + {x^2} \\ 1 + 1 + {x^2} + 2x = 1 + {x^2} \\ 1 + 2x = 0 \\ x = - \frac{1}{2} \\\end{aligned} \end{equation} $$

Question 12. Let $f:[0,4\pi ] \to [0,\pi ]$ be defined by $f(x) = {\cos ^{ - 1}}(\cos x)$. The number of points $x \in [0,4\pi ]$ satisfying the equation $f(x) = \frac{{10 - x}}{{10}}$ is

Solution: \[{\cos ^{ - 1}}(\cos x) = \left\{ \begin{gathered} x;x \in [0,\pi ] \hspace{1em} \\ 2\pi + x;x \in [\pi ,2\pi ] \hspace{1em} \\ - 2\pi + x;x \in [2\pi ,3\pi ] \hspace{1em} \\ 4\pi - x;x \in [3\pi ,4\pi ] \hspace{1em} \\ \end{gathered} \right\}\]

And $$\begin{equation} \begin{aligned} f(x) = \frac{{10 - x}}{{10}} \\ y = \frac{{10 - x}}{{10}} \\\end{aligned} \end{equation} $$$$y = 1 - \frac{x}{{10}}$$ The equation of line will pass through $(0,1)$ and $(10,0)$.

To find the number of solutions of equations involving trigonometric functions and algebraic functions we plot the graph and check the number of points of intersection as shown in figure.

As both graphs intersect at three points so a number of solution are $3$.