The correct option is (c) \[\mathbf{0}.\mathbf{96}\] We have, \[sin\text{ }(2\text{ }ta{{n}^{-1}}\left( 0.75 \right))\]
The correct option is (b) \[\mathbf{2}/\mathbf{5}\] Given, \[cos\text{ }(si{{n}^{-1}}~2/5\text{ }+\text{ }co{{s}^{-1}}~x)\text{ }=\text{ }0\] So, this can be rewritten as \[\begin{array}{*{35}{l}}...
The correct option is (a) \[\left[ \mathbf{1},\text{ }\mathbf{2} \right]\] We know that, \[si{{n}^{-1}}~x\] is defined for \[x\in \left[ -1,\text{ }1 \right]\] So, f(x) =...
The correct option is (a) \[\left[ \mathbf{0},\text{ }\mathbf{1} \right]\] Since, \[cos-1\text{ }x\] is defined for \[x\in \left[ -1,\text{ }1 \right]\] So, f(x) = \[\mathbf{cos}-\mathbf{1}\text{...
The correct option is (d) \[-\mathbf{\pi }/\mathbf{10}\]
The correct option is (b) \[\mathbf{1}\] Given, \[\mathbf{3}\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{x}\text{ }+\text{ }\mathbf{co}{{\mathbf{t}}^{-\mathbf{1}}}~\mathbf{x}\text{...
The correct option is (d) \[\left[ -\mathbf{\pi }/\mathbf{2},\text{ }\mathbf{\pi }/\mathbf{2} \right]\text{ }\text{ }\left\{ \mathbf{0} \right\}\] According to the principal branch of...
The correct answer is (c) \[\left[ \mathbf{0}.\text{ }\mathbf{\pi } \right]\] According to the principal value branch \[co{{s}^{-1}}~x\] is \[\left[ 0,\text{ }\pi \right]\].
Given \[{{\mathbf{a}}_{\mathbf{1}}},\text{ }{{\mathbf{a}}_{\mathbf{2}}},\text{ }{{\mathbf{a}}_{\mathbf{3}}},\text{ }\ldots .,\text{ }{{\mathbf{a}}_{\mathbf{n}}}~\] is an arithmetic progression with...
We have, \[ta{{n}^{-1}}\left( 1/2\text{ }si{{n}^{-1}}~3/4 \right)\] Let, \[~{\scriptscriptstyle 1\!/\!{ }_2}\text{ }si{{n}^{-1}}~{\scriptscriptstyle 3\!/\!{ }_4}\text{ }=\text{ }\theta \] or,...
\[\begin{array}{*{35}{l}} 4\text{ }ta{{n}^{-1}}~1/5\text{ }\text{ }ta{{n}^{-1}}~1/239 \\ =\text{ }2\text{ }(ta{{n}^{-1}}~1/5)\text{ }\text{ }ta{{n}^{-1}}~1/239 \\ \end{array}\] \[=\text{ }2\text{...
Taking the LHS, \[ta{{n}^{-1}}~1/4\text{ }+\text{ }ta{{n}^{-1}}~2/9\] = RHS – Hence Proved
Solution: Here, \[si{{n}^{-1}}~5/13\text{ }=\text{ }ta{{n}^{-1}}~5/12\] And, \[co{{s}^{-1}}~3/5\text{ }=\text{ }ta{{n}^{-1}}~4/3\] Taking the L.H.S, we have Thus, L.H.S = R.H.S – Hence Proved...
Taking the L.H.S, = \[\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{8}/\mathbf{17}\text{ }+\text{ }\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{3}/\mathbf{5}\] = tan-1 8/15 + tan-1 3/4 – Hence...
We have, \[\mathbf{co}{{\mathbf{s}}^{-\mathbf{1}}}~\left[ \mathbf{3}/\mathbf{5}\text{ }\mathbf{cos}\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{4}/\mathbf{5}\text{ }\mathbf{sin}\text{ }\mathbf{x}...
Taking L.H.S, = R.H.S – Hence Proved
Given equation, \[\mathbf{cos}\text{ }(\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}~}}\mathbf{x})\text{ }=\text{ }\mathbf{sin}\text{ }(\mathbf{co}{{\mathbf{t}}^{-\mathbf{1}~}}\mathbf{3}/\mathbf{4})\]...
Taking L.H.S, we have Thus, L.H.S = R.H.S – Hence proved
Given, \[\mathbf{2}\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}\left( \mathbf{cos}\text{ }\mathbf{\theta } \right)\text{ }=\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\left( \mathbf{2}\text{...
Given expression, \[\mathbf{sin}\text{ }(\mathbf{2}\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\mathbf{1}/\mathbf{3})\text{ }+\text{ }\mathbf{cos}\text{...
Given equation, Hence, the real solutions of the given trigonometric equation are \[0\] and \[-1\].
Taking L.H.S = \[2\text{ }ta{{n}^{-1}}\left( -3 \right)\text{ }=\text{ }-2\text{ }ta{{n}^{-1}}~3\text{ }(\because ta{{n}^{-1}}~\left( -x \right)\text{ }=\text{ }\text{ }ta{{n}^{-1}}~x)\text{ }\in...
We know that, \[ta{{n}^{-1}}~tan\text{ }x\text{ }=\text{ }x,\text{ }x\in \left( -\pi /2,\text{ }\pi /2 \right)\]
According to the question, \[\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}~\left( -\mathbf{1}/\surd \mathbf{3} \right)\text{ }+\text{ }\mathbf{co}{{\mathbf{t}}^{-\mathbf{1}}}\left( \mathbf{1}/\surd...
According to the question, L.H.S = R.H.S – Hence Proved
We know that, \[ta{{n}^{-1}}~tan\text{ }x\text{ }=\text{ }x,\text{ }x\in \left( -\pi /2,\text{ }\pi /2 \right)\] And, here \[ta{{n}^{-1}}~tan\text{ }\left( 5\pi /6 \right)\text{ }=\text{ }5\pi...