JEE MAIN - Mathematics (2021 - 26th February Morning Shift - No. 11)
If $${{{{\sin }^1}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\tan }^{ - 1}}y} \over c}$$; $$0 < x < 1$$,
then the value of $$\cos \left( {{{\pi c} \over {a + b}}} \right)$$ is :
then the value of $$\cos \left( {{{\pi c} \over {a + b}}} \right)$$ is :
$${{1 - {y^2}} \over {2y}}$$
$${{1 - {y^2}} \over {y\sqrt y }}$$
$$1 - {y^2}$$
$${{1 - {y^2}} \over {1 + {y^2}}}$$
Explanation
$${{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\tan }^{ - 1}}y} \over c}$$
$${{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \over {a + b}} = {\pi \over {2(a + b)}}$$
Now, $${{{{\tan }^{ - 1}}y} \over c} = {\pi \over {2(a + b)}}$$
$$2{\tan ^{ - 1}}y = {{\pi c} \over {a + b}}$$
$$ \Rightarrow \cos \left( {{{\pi c} \over {a + b}}} \right) = \cos (2{\tan ^{ - 1}}y) = {{1 - {y^2}} \over {1 + {y^2}}}$$
$${{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \over {a + b}} = {\pi \over {2(a + b)}}$$
Now, $${{{{\tan }^{ - 1}}y} \over c} = {\pi \over {2(a + b)}}$$
$$2{\tan ^{ - 1}}y = {{\pi c} \over {a + b}}$$
$$ \Rightarrow \cos \left( {{{\pi c} \over {a + b}}} \right) = \cos (2{\tan ^{ - 1}}y) = {{1 - {y^2}} \over {1 + {y^2}}}$$
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