Solve the following equations: `sin^(-1)(3x)/5+sin^(-1)(4x)/5=sin^(-1)x` `sin^(-1)6x+sin^(-1)6sqrt(3)x=pi/2`

Explanation

We have,` sin^(−1)(3x/5)​+sin^(−1)(4x/5)​=sin^(−1)x`


⇒`sin^(−1){(3x/5)​sqrt(1−(16x^2)/25)​
​+(4x/5)​sqrt(1−(9x^2)/25)​
​}=sin^(−1)x`

⇒`3xsqrt(25−16x^2)
​+4xsqrt(25−9x^2)
​=25x`

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