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Question: How do you find \[\cot \left( {{{\sin }^{ - 1}}\left( {\dfrac{p}{q}} \right)} \right)?\]...

How do you find cot⁡(sin⁡−1(pq))?\cot \left( {{{\sin }^{ - 1}}\left( {\dfrac{p}{q}} \right)} \right)?

Explanation

Solution

In the given problem first we need to concentrate on different trigonometric ratios given in the problem, here it is cotangent and sine. We are trying to convert sine inverse into cotangent inverse so that they will get canceled out and the required value will get obtained.

Formula used:
Pythagoras theorem, h2=p2+b2{h^2} = {p^2} + {b^2} where, (h=h = Hypotenuse, p=p = perpendicular, b=b = base)
Cosec\theta $$$$ = \left( {\dfrac{h}{p}} \right) where, (h=h = Hypotenuse, p=p = perpendicular)

Complete step by step solution:
First try to convert sine inverse into cot inverse, for that let sine inverse be ‘θ\theta ’ and write it as follows
⇒θ=sin⁡−1(pq)\Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{p}{q}} \right)
Taking sine inverse to left hand side, we get
⇒sin⁡θ=(pq)\Rightarrow \sin \theta = \left( {\dfrac{p}{q}} \right)
Comparing above equation with sin⁡θ=ph\sin \theta = \dfrac{p}{h} where, (h=h = Hypotenuse, p=p = perpendicular), we get
⇒p=p\Rightarrow p = p and h=qh = q
Therefore, we got following equation,
\Rightarrow $$$$\cos ec\theta = \left( {\dfrac{q}{p}} \right)
So, we can replace sin⁡θ\sin \theta with cos⁡ecθ\cos ec\theta and write it as following,
\Rightarrow $$$$\cos ec\theta = \left( {\dfrac{q}{p}} \right)
Taking cos⁡ec\cos ec to the right-hand side so that it will get converted into inverse form,
\Rightarrow $$$$\theta = \cos e{c^{ - 1}}\left( {\dfrac{q}{p}} \right)
Now, the expression given in problem can be replaced by above equation, we get
\Rightarrow $$$$\cot \left( {\cos e{c^{ - 1}}\left( {\dfrac{q}{p}} \right)} \right)
Now, we are trying to convert the above expression into cotangent inverse form.
For doing so, we have to calculate cot⁡θ\cot \theta in the form of qq and pp. cot⁡θ\cot \theta can be written as,
⇒cot⁡θ=bp\Rightarrow \cot \theta = \dfrac{b}{p} where, (b=b = base, p=p = perpendicular)
By using Pythagoras theorem, it can also be written as following,
⇒cot⁡θ=h2−p2p\Rightarrow \cot \theta = \dfrac{{\sqrt {{h^2} - {p^2}} }}{p}
As we know by above conclusion that p=pp = p and h=qh = q, it can also be written as,
⇒cot⁡θ=q2−p2p\Rightarrow \cot \theta = \dfrac{{\sqrt {{q^2} - {p^2}} }}{p}
Taking pp common from numerator, we get
⇒cot⁡θ=p2(q2p2−1)p\Rightarrow \cot \theta = \dfrac{{\sqrt {{p^2}\left( {\dfrac{{{q^2}}}{{{p^2}}} - 1} \right)} }}{p}
Taking pp outside from under-root numerator, we get
⇒cot⁡θ=p(q2p2−1)p\Rightarrow \cot \theta = \dfrac{{p\sqrt {\left( {\dfrac{{{q^2}}}{{{p^2}}} - 1} \right)} }}{p}
pp will get cancel out with denominator,
⇒cot⁡θ=(q2p2−1)\Rightarrow \cot \theta = \sqrt {\left( {\dfrac{{{q^2}}}{{{p^2}}} - 1} \right)}
Simplifying above equation by taking L.C.M, we get
⇒cot⁡θ=(q2−p2p2)\Rightarrow \cot \theta = \sqrt {\left( {\dfrac{{{q^2} - {p^2}}}{{{p^2}}}} \right)}
Taking pp outside from under-root denominator, we get
⇒cot⁡θ=q2−p2p\Rightarrow \cot \theta = \dfrac{{\sqrt {{q^2} - {p^2}} }}{p}
Taking cot⁡\cot right hand side so that it will get converted into inverse form as follows,
⇒θ=cot⁡−1q2−p2p\Rightarrow \theta = {\cot ^{ - 1}}\dfrac{{\sqrt {{q^2} - {p^2}} }}{p}
It can also be written as following,
\Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{p}{q}} \right) = $$$${\cot ^{ - 1}}\dfrac{{\sqrt {{q^2} - {p^2}} }}{p}
So, we are going to replace the expression given in problem with the equation obtained above,
\Rightarrow $$$$\cot \left( {{{\cot }^{ - 1}}\dfrac{{\sqrt {{q^2} - {p^2}} }}{p}} \right)
cot⁡\cot will get cancel out and we obtain the final required value as following,
\Rightarrow $$$$\left( {\dfrac{{\sqrt {{q^2} - {p^2}} }}{p}} \right)

Note:
We have to make relations between different trigonometric ratios so that one trigonometric ratio can be easily converted into another trigonometric ratio with the help of Pythagoras theorem.