"If $${{{{\sin }^1}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\tan }^{ - 1}}y} \over c}$$; $0

Question

"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 :"

Accepted Answer

"$${{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\tan }^{ - 1}}y} \over c}$$

$${{{{\sin }^{ - 1}}x} \over a} = {{{{\cos }^{ - 1}}x} \over b} = {{{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \over {a + b}} = {\pi \over {2(a + b)}}$$

Now, ${{{{\tan }^{ - 1}}y} \over c} = {\pi \over {2(a + b)}}$

$2{\tan ^{ - 1}}y = {{\pi c} \over {a + b}}$

$$ \Rightarrow \cos \left( {{{\pi c} \over {a + b}}} \right) = \cos (2{\tan ^{ - 1}}y) = {{1 - {y^2}} \over {1 + {y^2}}}$$"

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 :

Solution

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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 :

Solution

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More Inverse Trigonometric Functions questions

Drill 25 more like these. Every day. Free.

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 :