Question

Prove that cosec\left[ {{{\tan }^{ - 1}}\left\\{ {\cos \left\\{ {{{\cot }^{ - 1}}\left\\{ {\sec \left( {{{\sin }^{ - 1}}a} \right)} \right\\}} \right\\}} \right\\}} \right] = \sqrt {3 - {a^2}} where $0 < a < 1$.

Answer 1

In this question, we have to verify the given trigonometric equation of inverse trigonometric ratio. We will first proceed with LHS and solve it. We proceed by assuming ${\sin ^{ - 1}}a = x$ and then take sine both side and use the fact $\sin ({\sin ^{ - 1}}a) = a$ to get $a = \sin x$. Now we find the value of other T-ratio as required and put it back into LHS. We will proceed in a similar manner to get the required RHS.

Complete answer:
The given question is based on the inverse trigonometry ratio of angles. The inverse trigonometry ratio is the inverse of trigonometric ratio sine, cosine, tangent, cosecant, secant and cotangent. For example: If $sin{\text{ }}a{\text{ }} = {\text{ }}x$ then the inverse trigonometric ratio is written as ${\sin ^{ - 1}}x = a$.
Consider the given question,
LHS = cosec\left[ {{{\tan }^{ - 1}}\left\\{ {\cos \left\\{ {{{\cot }^{ - 1}}\left\\{ {\sec \left( {{{\sin }^{ - 1}}a} \right)} \right\\}} \right\\}} \right\\}} \right]
Let us consider, ${\sin ^{ - 1}}a = x$
Taking sine both side we get,
$\sin ({\sin ^{ - 1}}a) = \sin x$
We know that $\sin ({\sin ^{ - 1}}a) = a$
$\therefore a = \sin x$
We find the value of $\sec x$ using $sin{\text{ }}a{\text{ }} = {\text{ }}x$.
Hence , $\sec x = \dfrac{1}{{\sqrt {1 - {a^2}} }}$
$\Rightarrow x = {\sec ^{ - 1}}\dfrac{1}{{\sqrt {1 - {a^2}} }}$
Putting ${\sin ^{ - 1}}a = x = {\sec ^{ - 1}}\dfrac{1}{{\sqrt {1 - {a^2}} }}$ in LHS. we have,
LHS = cosec\left[ {{{\tan }^{ - 1}}\left\\{ {\cos \left\\{ {{{\cot }^{ - 1}}\left\\{ {\sec \left( {{{\sec }^{ - 1}}\dfrac{1}{{\sqrt {1 - {a^2}} }}} \right)} \right\\}} \right\\}} \right\\}} \right]
Now we know that the value of $\sec \left( {{{\sec }^{ - 1}}\dfrac{1}{{\sqrt {1 - {a^2}} }}} \right) = \dfrac{1}{{\sqrt {1 - {a^2}} }}$
From LHS, we have
LHS = cosec\left[ {{{\tan }^{ - 1}}\left\\{ {\cos \left\\{ {{{\cot }^{ - 1}}\left\\{ {\dfrac{1}{{\sqrt {1 - {a^2}} }}} \right\\}} \right\\}} \right\\}} \right]
Now, consider y = {\cot ^{ - 1}}\left\\{ {\dfrac{1}{{\sqrt {1 - {a^2}} }}} \right\\}
Again Talking $\cot$ both side we have
\cot y = \cot \left( {{{\cot }^{ - 1}}\left\\{ {\dfrac{1}{{\sqrt {1 - {a^2}} }}} \right\\}} \right)
We know that, \cot \left\\{ {{{\cot }^{ - 1}}\left\\{ {\dfrac{1}{{\sqrt {1 - {a^2}} }}} \right\\}} \right\\} = \dfrac{1}{{\sqrt {1 - {a^2}} }}
Hence $\cot y = \dfrac{1}{{\sqrt {1 - {a^2}} }}$
Using $\cot y = \dfrac{1}{{\sqrt {1 - {a^2}} }}$ to find the value of cosine, we have
$\cos y = \dfrac{1}{{\sqrt {2 - {a^2}} }}$
$\Rightarrow y = {\cos ^{ - 1}}\dfrac{1}{{\sqrt {2 - {a^2}} }}$
Putting {\cot ^{ - 1}}\left\\{ {\dfrac{1}{{\sqrt {1 - {a^2}} }}} \right\\} = y = {\cos ^{ - 1}}\dfrac{1}{{\sqrt {2 - {a^2}} }} in LHS we have,
LHS = cosec\left[ {{{\tan }^{ - 1}}\left\\{ {\cos \left\\{ {{{\cos }^{ - 1}}\dfrac{1}{{\sqrt {2 - {a^2}} }}} \right\\}} \right\\}} \right]
We know that, \cos \left\\{ {{{\cos }^{ - 1}}\dfrac{1}{{\sqrt {2 - {a^2}} }}} \right\\} = \dfrac{1}{{\sqrt {2 - {a^2}} }}, we have
LHS = cosec\left[ {{{\tan }^{ - 1}}\left\\{ {\dfrac{1}{{\sqrt {2 - {a^2}} }}} \right\\}} \right]
Again consider, z = {\tan ^{ - 1}}\left\\{ {\dfrac{1}{{\sqrt {2 - {a^2}} }}} \right\\}
Taking tangent both side we have,
\tan z = \tan \left\\{ {{{\tan }^{ - 1}}\left\\{ {\dfrac{1}{{\sqrt {2 - {a^2}} }}} \right\\}} \right\\} = \dfrac{1}{{\sqrt {2 - {a^2}} }}
Therefore, $\cos ec{\kern 1pt} {\kern 1pt} z = \sqrt {3 - {a^2}}$
$\Rightarrow z = \cos e{c^{ - 1}}\sqrt {3 - {a^2}}$
Putting {\tan ^{ - 1}}\left\\{ {\dfrac{1}{{\sqrt {2 - {a^2}} }}} \right\\} = z = \cos e{c^{ - 1}}\sqrt {3 - {a^2}} in LHS we have
LHS = cosec\left[ {{{\tan }^{ - 1}}\left\\{ {\dfrac{1}{{\sqrt {2 - {a^2}} }}} \right\\}} \right]
LHS$= cosec\left[ {\cos e{c^{ - 1}}\sqrt {3 - {a^2}} } \right]$
On solving, we have
LHS$= \sqrt {3 - {a^2}}$
Therefore, LHS $=$RHS
Hence proved. cosec\left[ {{{\tan }^{ - 1}}\left\\{ {\cos \left\\{ {{{\cot }^{ - 1}}\left\\{ {\sec \left( {{{\sin }^{ - 1}}a} \right)} \right\\}} \right\\}} \right\\}} \right] = \sqrt {3 - {a^2}}

Note:
To find the value of any T-Ratio from any T-Ratio we use Pythagorean Theorem.
For example, Let $sin{\text{ x }} = a{\text{ }}$ and we have to find the value of $\tan x$.
We proceed as follow:
$\sin x = \dfrac{p}{h} = \dfrac{x}{1}$ , Where $p$ is perpendicular and $h$ is hypotenuse.
Hence $p = x$ and $h = 1$.
From Pythagorean Theorem, we have
$b = \sqrt {{h^2} - {p^2}}$ Where $b$ is base .
Hence $b = \sqrt {{1^2} - {x^2}} = \sqrt {1 - {x^2}}$
Now we find the value of tangent as follow,
$\tan x = \dfrac{p}{b} = \dfrac{x}{{\sqrt {1 - {x^2}} }}$.
Similarly we can find any other T-Ratio.