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Prove the given inverse trigonometric expression,9π894sin113=94sin1223.

Answer
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Hint: In this question, we can see that 94 is common in both of the terms present in LHS of the expression. So, we can take 94 as common. Using the property sin1θ+cos1θ=π2 , we can simplify the expression and then transform the inverse cosine function into inverse sine function.

Complete step-by-step solution -
Let us first consider the LHS of the given expression. Now taking 94 as common, we get
9π894sin113=94(π2sin113)
In the above expression, we have (π2sin113) . So, we need to convert this into a simpler form.
And we also know that the property that, sin1θ+cos1θ=π2…………(1)
In the above equation (1), taking sin1θ to the RHS, we get
cos1θ=π2sin1θ………….(2)
We are replacing θ by 13 in the equation (2), we get
cos113=π2sin113……………..(3)
Now, according to the question we have,
94(π2sin113)…………….(4)
Using equation (3), and putting the value of π2sin113 in equation (4). We get,
94(π2sin113)
=94cos113………………..(5)
But according to the question, we have the inverse of sine in RHS. So, here we have to convert it into the inverse of sine.
Converting inverse cosine function to inverse sine function.
Let us assume,
x=cos113……………..(6)
Taking cosine in both of LHS as well as RHS in the equation (6), we get
x=cos113cosx=13
Now, we have to find sinx .
For that, we also know the identity
sin2x+cos2x=1sin2x=1cos2xsinx=1cos2x
Using this formula, we can get the value of sinx .
sinx=1cos2xsinx=119sinx=919sinx=89sinx=223
Now we have, sinx=223……………(7)
Taking the inverse of sine in both LHS as well as RHS in equation (7), we get
sinx=223
x=sin1223…………….(8)
From equation (5), we have 94cos113 .
Using equation (6), we can write the equation (5) as 94x .
And by using equation (8) we can write 94x as, 94sin1223………(9)
So, 9π894sin113=94sin1223
Therefore, LHS = RHS.
Hence, proved.

Note: In this question, one can think to find the principal value of inverse sine functions which are provided in LHS and RHS of the given equation, that is sin113 and 94sin1223 . If we do so, then only we are increasing the complexity and here, we don’t need the principal value of inverse sine function.

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