We often need to convert inverse of sin to either inverse of cos, tan, sec, cosec , cot .In this post we will see how we can do it easily

## Inverse of sin to inverse of cos

**Case 1**

$sin^{-1} x$ and x > 0

Now we can write as

$\theta=sin^{-1} x$

$sin \theta =x$

Now we know that here $\theta \in [0,\pi/2]$, so it is an acute angle

Now it can be written as

$sin \theta =\frac {x}{1} = \frac {perp}{hyp}$

In Right angle triangle

Now then base becomes

$\text{Base} = \sqrt { 1 -x^2}$

So , $cos \theta = \frac {base}{hyp} = \sqrt { 1 -x^2}$

or

$\theta= cos^{-1} \sqrt { 1 -x^2}$

or $sin^{-1} x = cos^{-1} \sqrt { 1 -x^2}$

**Case **II

$sin^{-1} x$ and x < 0

So value of the function will be in the range $[-\pi/2 , 0]$

Now we know from the property that

$sin^{-1} (-x)= – sin^{-1} (x)$

This can be written as

$sin^{-1} x = – sin^{-1} |x| = – cos^{-1} \sqrt { 1 -|x|^2}= – cos^{-1} \sqrt { 1 -x^2}$

This makes sense also as Range of the cos and sin function differ. We can convert with out worrying about the sign in $[0, \pi/2] as it is common

Thus ,we have different formula depending on the values of x

## Inverse of sin to inverse of tan

**Case 1**

$sin^{-1} x$ and x > 0

from the above, we can write that

$tan \theta = \frac {perp}{base} = \frac {x}{\sqrt {1-x^2}}$

or

$sin^{-1} x = tan^{-1} \frac {x}{\sqrt { 1 -x^2}}$

**Case **II

$sin^{-1} x$ and x < 0

Now we know from the property that

$sin^{-1} (-x)= – sin^{-1} (x)$

This can be written as

$sin^{-1} x = – sin^{-1} |x| = – tan^{-1} \frac {x}{\sqrt { 1 -x^2}}= tan^{-1} \frac {-|x|}{\sqrt { 1 -x^2}} = tan^{-1} \frac {x}{\sqrt { 1 -x^2}}$

So we have same formula for any values of x

$sin^{-1} x = tan^{-1} \frac {x}{\sqrt { 1 -x^2}}$

## Inverse of sin to inverse of cosec

This we already know from the property

$sin^{-1} x = cosec^{-1} \frac {1}{x}$

for all values of x [-1,1]

## Inverse of sin to inverse of sec

**Case 1**

$sin^{-1} x$ and x > 0

from the above, we can write that

$sec \theta = \frac {hyp}{base} = \frac {1}{\sqrt {1-x^2}}$

or

$sin^{-1} x = sec^{-1} \frac {1}{\sqrt { 1 -x^2}}$

**Case **II

$sin^{-1} x$ and x < 0

Now we know from the property that

$sin^{-1} (-x)= – sin^{-1} (x)$

This can be written as

$sin^{-1} x = – sin^{-1} |x| = – sec^{-1} \frac {1}{\sqrt { 1 -x^2}} = – sec^{-1} \frac {1}{\sqrt { 1 -x^2}}$

This makes sense also as Range of the sec and sin function differ. We can convert with out worrying about the sign in $[0, \pi/2] as it is common

Thus ,we have different formula depending on the values of x

## Inverse of sin to inverse of cot

**Case 1**

$sin^{-1} x$ and x > 0

from the above, we can write that

$cot \theta = \frac {base}{perp} = \frac{\sqrt {1-x^2}}{x}$

or

$sin^{-1} x = cot^{-1} \frac {\sqrt { 1 -x^2}}{x}$

**Case **II

$sin^{-1} x$ and x < 0

Now we know from the property that

$sin^{-1} (-x)= – sin^{-1} (x)$

This can be written as

$sin^{-1} x = – sin^{-1} |x| = – cot^{-1} \frac {\sqrt { 1 -x^2}}{|x|}$

This makes sense also as Range of the sec and sin function differ. We can convert with out worrying about the sign in $[0, \pi/2] as it is common

Thus ,we have different formula depending on the values of x