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Mathematics Question on Trigonometric Equations

If asin−1x−bcos−1x=casin^{-1}x - bcos^{-1} x = c , asin−1x+bcos−1xa sin^{-1}x + b cos^{-1} x is equal to

A

πab+c(a−b)a+b\frac{\pi ab +c(a-b)}{a+b}

B

00

C

πab−c(a−b)a+b\frac {\pi ab -c (a-b)}{a+b}

D

π2\frac{\pi}{2}

Answer

πab+c(a−b)a+b\frac{\pi ab +c(a-b)}{a+b}

Explanation

Solution

Given,
a sin−1 x−b cos−1 x=c   ...(i)a\, sin^{-1} \,x - b \,cos^{-1} \,x = c\,\,\,...(i)
⇒a(π2−cos−1x)−b cos−1x=c\Rightarrow a\left(\frac{\pi}{2} - cos^{-1}x\right) - b\, cos^{-1} x=c
⇒(a+b)cos−1x=aπ2−c\Rightarrow \left(a + b\right)cos^{-1}x = \frac{a\pi}{2} - c
⇒cos−1x=aπ/2−ca+b\Rightarrow cos^{-1} x = \frac{a\pi /2 -c}{a+b}
Again from E (i)(i)
a sin−1x−b(π2−sin−1 x)=ca\, sin^{-1} x - b (\frac{\pi}{2} - sin^{-1} \,x ) = c
⇒(sin−1x)−(a+b)=c+bπ2\Rightarrow (sin^{-1} x) - ( a + b ) = c + \frac{b\pi}{2}
⇒sin−1x=c+bπ/2a+b\Rightarrow sin^{-1} x = \frac{c + b\pi/2}{a + b}
∴a sin−1 x+b cos−1 x=a(c+bπ/2)a+b\therefore a\, sin^{-1} \,x + b \,cos^{-1}\,x = \frac{a(c + b\pi/2)}{a + b}
+b(aπ/2−c)a+b+ \frac{b(a\pi/2 - c )}{a + b}
=c(a−b)+abπa+b= \frac{c(a - b) + ab\pi}{a + b}