Inverse Trigonometry Function MCQ Questions & Answers in Trigonometry | Maths

$$\eqalign{ & {\text{Let, }}{\cos ^{ - 1}}\sqrt p + {\cos ^{ - 1}}\sqrt {1 - p} + {\cos ^{ - 1}}\sqrt {1 - q} = \frac{{3\pi }}{4}\,\,.....\left( {\text{i}} \right) \cr & {\text{Let, }}\,a = {\cos ^{ - 1}}\sqrt p \,b = {\cos ^{ - 1}}\sqrt {1 - p} {\text{ and }}c = {\cos ^{ - 1}}\sqrt {1 - q} \cr & \Rightarrow \cos a = \sqrt p ,\cos b = \sqrt {1 - p} ,\cos c = \sqrt {1 - q} \cr & \Rightarrow {\cos ^2}a = p,{\cos ^2}b = 1 - p,{\cos ^2}c = 1 - q \cr & {\text{Now}},\,\,{\sin ^2}a = 1 - {\cos ^2}a = 1 - p \cr & \Rightarrow \sin a = \sqrt {1 - p} , \cr & {\sin ^2}b = 1 - {\cos ^2}b = 1 - 1 + p \cr & \Rightarrow \sin b = \sqrt p \cr & {\sin ^2}c = 1 - {\cos ^2}c = 1 - 1 + q = q \cr & \Rightarrow \sin c = \sqrt q \cr} $$
$$\therefore $$ equation (i) can be written as
$$\eqalign{ & a + b + c = \frac{{3\pi }}{4} \cr & \Rightarrow a + b = \frac{{3\pi }}{4} - c \cr} $$
Take $$\cos$$  on each side, we get
$$\eqalign{ & \cos \left( {a + b} \right) = \cos \left( {\frac{{3\pi }}{4} - c} \right) \cr & \Rightarrow \cos a\cos b - \sin a\sin b \cr & = \cos \left\{ {\pi - \left( {\frac{\pi }{4} + c} \right)} \right\} = - \cos \left( {\frac{\pi }{4} + c} \right) \cr} $$
Put values of $$\cos a, \cos b$$   and $$\sin a, \sin b,$$   we get
$$\eqalign{ & \sqrt p \cdot \sqrt {1 - p} - \sqrt {1 - p} \sqrt p \cr & = - \left( {\frac{1}{{\sqrt 2 }}\sqrt {1 - q} - \frac{1}{{\sqrt 2 }}\sqrt q } \right) \cr & \Rightarrow 0 = \sqrt {1 - q} - \sqrt q \cr & \Rightarrow \sqrt {1 - q} = \sqrt q \cr} $$
Squaring on both side : $$ \Rightarrow 1 - q = q$$
$$\eqalign{ & \Rightarrow 1 = 2q \cr & \Rightarrow q = \frac{1}{2} \cr} $$

$$\eqalign{ & {\left[ {{{\cot }^{ - 1}}x} \right]^2} - 6\left[ {{{\cot }^{ - 1}}x} \right] + 9 \leqslant 0 \cr & \Rightarrow {\left( {\left[ {{{\cot }^{ - 1}}x} \right] - 3} \right)^2} \leqslant 0 \cr & \Rightarrow \left[ {{{\cot }^{ - 1}}x} \right] = 3 \cr & \Rightarrow 3 \leqslant {\cot ^{ - 1}}x < 4 \cr & \Rightarrow x \in \left( { - \infty ,\cot 3} \right] \cr} $$

We have,
$$\eqalign{ & {\tan ^{ - 1}}\left( {\frac{d}{{1 + {a_1}{a_2}}}} \right) + {\tan ^{ - 1}}\left( {\frac{d}{{1 + {a_2}{a_3}}}} \right) + ..... + {\tan ^{ - 1}}\left( {\frac{d}{{1 + {a_{n - 1}}{a_n}}}} \right) \cr & = {\tan ^{ - 1}}\left( {\frac{{{a_2} - {a_1}}}{{1 + {a_1}{a_2}}}} \right) + {\tan ^{ - 1}}\left( {\frac{{{a_3} - {a_2}}}{{1 + {a_2}{a_3}}}} \right) + ..... + {\tan ^{ - 1}}\left( {\frac{{{a_n} - {a_{n - 1}}}}{{1 + {a_{n - 1}}{a_n}}}} \right) \cr & = \left( {{{\tan }^{ - 1}}{a_2} - {{\tan }^{ - 1}}{a_1}} \right) + \left( {{{\tan }^{ - 1}}{a_3} - {{\tan }^{ - 1}}{a_2}} \right) + ..... + \left( {{{\tan }^{ - 1}}{a_n} - {{\tan }^{ - 1}}{a_{n - 1}}} \right) \cr & = {\tan ^{ - 1}}{a_n} - {\tan ^{ - 1}}{a_1} = {\tan ^{ - 1}}\left( {\frac{{{a_n} - {a_1}}}{{1 + {a_n}{a_1}}}} \right) \cr & = {\tan ^{ - 1}}\left( {\frac{{\left( {n - 1} \right)d}}{{1 + {a_1}{a_n}}}} \right) \cr & \therefore \tan \left[ {{{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_1}{a_2}}}} \right) + {{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_2}{a_3}}}} \right) + ..... + {{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_{n - 1}}{a_n}}}} \right)} \right] = \frac{{\left( {n - 1} \right)d}}{{1 + {a_1}{a_n}}} \cr} $$

$$\eqalign{ & {\cot ^{ - 1}}\left( {1 + \sum\limits_{k = 1}^n {2k} } \right) \cr & = {\cot ^{ - 1}}\left[ {1 + n\left( {n + 1} \right)} \right] \cr & = {\tan ^{ - 1}}\left[ {\frac{{\left( {n + 1} \right) - n}}{{1 + \left( {n + 1} \right)n}}} \right] \cr & = {\tan ^{ - 1}}\left( {n + 1} \right) - {\tan ^{ - 1}}n \cr & \therefore \,\,\sum\limits_{n = 1}^{23} {\left[ {{{\tan }^{ - 1}}\left( {n + 1} \right) - {{\tan }^{ - 1}}n} \right]} \cr & = {\tan ^{ - 1}}24 - {\tan ^{ - 1}}1 \cr & = {\tan ^{ - 1}}\frac{{23}}{{25}} \cr & \therefore \,\,\cot \left[ {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{k = 1}^n {2k} } \right)} } \right] \cr & = \cot \left[ {{{\tan }^{ - 1}}\frac{{23}}{{25}}} \right] \cr & = \frac{{25}}{{23}} \cr} $$

$$\eqalign{ & \tan \left[ {{{\cos }^{ - 1}}\frac{1}{{5\sqrt 2 }} - {{\sin }^{ - 1}}\frac{4}{{\sqrt {17} }}} \right] \cr & = \tan \left[ {{{\tan }^{ - 1}}7 - {{\tan }^{ - 1}}4} \right] \cr & = \tan \left( {{{\tan }^{ - 1}}\left( {\frac{3}{{29}}} \right)} \right) \cr & = \frac{3}{{29}} \cr} $$

$${\tan ^{ - 1}} + {\cos ^{ - 1}}\left( {\frac{y}{{1 + {y^2}}}} \right) = {\sin ^{ - 1}}\left( {\frac{3}{{\sqrt {10} }}} \right)$$
We will convert this equation in $${\tan ^{ - 1}}$$  from
$$\eqalign{ & \Rightarrow {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {\frac{1}{y}} \right) = {\tan ^{ - 1}}3 \cr & \Rightarrow {\tan ^{ - 1}}\left( {\frac{1}{y}} \right) = {\tan ^{ - 1}}3 - {\tan ^{ - 1}}x \cr & \Rightarrow {\tan ^{ - 1}}\left( {\frac{1}{y}} \right) = {\tan ^{ - 1}}\left( {\frac{{3 - x}}{{1 + 3x}}} \right) \cr & \Rightarrow \frac{1}{y} = \frac{{3 - x}}{{1 + 3x}} \cr & \Rightarrow y = \frac{{1 + 3x}}{{3 - x}} \cr} $$
As, we have to find positive integral values, so $$x$$ can have only two valus 1 and 2.
$$\eqalign{ & {\text{When }}x = 1,\,y = \frac{{1 + 3}}{2} \Rightarrow y = 2 \cr & {\text{When }}x = 2,\,y = \frac{{1 + 6}}{1} \Rightarrow y = 7 \cr} $$
So, these are two possible solutions for the given equation.

The given relation is possible when
$$\eqalign{ & a - \frac{{{a^2}}}{3} + \frac{{{a^3}}}{9} + ..... = 1 + b + {b^2} + ..... \cr & {\text{Also }} - 1 \leqslant a - \frac{{{a^2}}}{3} + \frac{{{a^3}}}{9} + ..... \leqslant 1 \cr & {\text{and}}\, - 1 \leqslant 1 + b + {b^2} + ..... \leqslant 1 \cr & \Rightarrow \left| b \right| < 1 \cr & \Rightarrow \left| {a} \right| < 3 \,{\text{ and }}\,\frac{a}{{1 + \frac{a}{3}}} = \frac{1}{{1 - b}} \cr & \Rightarrow \frac{{3a}}{{a + 3}} = \frac{1}{{1 - b}} \cr} $$
There are infinitely many solutions. But in the given options, it is satisified only when
$$a = 1\,\,{\text{and }}\,b = - \frac{1}{3}.$$

$$ - \frac{\pi }{2} \leqslant {\sin ^{ - 1}}x \leqslant \frac{\pi }{2}.$$     Hence, from the question, $${\sin^{ - 1}}{x_i} = \frac{\pi }{2}$$   for all $$i.$$
$$\therefore {x_i} = 1$$  for all $$i.$$

$$\eqalign{ & \tan \left( {{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y + {{\tan }^{ - 1}}z} \right) - \cot \left( {{{\cot }^{ - 1}}x + {{\cot }^{ - 1}}y + {{\cot }^{ - 1}}z} \right) \cr & = \tan \left( {{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y + {{\tan }^{ - 1}}z} \right) - \cot \left( {\frac{\pi }{2} - {{\tan }^{ - 1}}x + \frac{\pi }{2} - {{\tan }^{ - 1}}y + \frac{\pi }{2} - {{\tan }^{ - 1}}z} \right)\,\left( {\because {{\tan }^{ - 1}}x + {{\cot }^{ - 1}}x = \frac{\pi }{2}} \right) \cr & = \tan \left( {{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y + {{\tan }^{ - 1}}z} \right) - \cot \left\{ {\frac{{3\pi }}{2} - \left( {{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y + {{\tan }^{ - 1}}z} \right)} \right. \cr & = \tan \left( {{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y + {{\tan }^{ - 1}}z} \right) - \tan \left( {{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y + {{\tan }^{ - 1}}z} \right) = 0 \cr} $$

$$\eqalign{ & {\left( {{{\sin }^{ - 1}}x} \right)^3} + {\left( {{{\cos }^{ - 1}}x} \right)^3} = \frac{\pi }{2}\left[ {{{\left( {\frac{\pi }{2}} \right)}^2} - 3\,{{\sin }^{ - 1}}x\,{{\cos }^{ - 1}}x} \right] \cr & = \frac{{3\pi }}{2}\left[ {{{\left( {{{\sin }^{ - 1}}x - \frac{\pi }{4}} \right)}^2} + \frac{{{\pi ^2}}}{{48}}} \right] \cr} $$