Question

value of sin(sin^-1 1/7 +tan^-1 1/4) is​

Answer 1

Step-by-step explanation:

Solution: (1) π / 4

sin-1 (3 / 5) + tan-1 (1 / 7) = tan-1 (3 / 4) + tan-1 (1 / 7)

= tan-1 [(3 / 4) + (1 / 7)] / [1 – (3 / 4) (1 / 7)]

= tan-1 (25 / 25)

= tan-1 1

= π / 4

Answer 2

Answer:

$\frac{4( 1 + \sqrt{3}) }{7\sqrt{17} }$

Step-by-step explanation:

Given :- The given expression is sin( $sin^{-1}$ ( $\frac{1}{7}$ ) + $tan^{-1}$ ( $\frac{1}{4}$ ) )

To Find:- The value of expression sin( $sin^{-1}$ ( $\frac{1}{7}$ ) + $tan^{-1}$ ( $\frac{1}{4}$ ) )

Solution :-

We know that, tan C = Perpendicular / Base

                Hypotenuse = √( Perpendicular² + Base² )

                $sin^{-1}$ x + $sin^{-1}$ y =  $sin^{-1}$ [ x √(1-y²) + y √(1-x²) ]

The given expression is sin( $sin^{-1}$ ( $\frac{1}{7}$ ) + $tan^{-1}$ ( $\frac{1}{4}$ ) )

We can write $tan^{-1}$ ( 1/4 ) = $sin^{-1}$ ( 1/√17 )

Now, the expression becomes

= sin( $sin^{-1}$ ( $\frac{1}{7}$ ) + $sin^{-1}$ ( $\frac{1}{\sqrt{17} }$ ) )

= sin( $sin^{-1}$ [ $\frac{1}{7} \sqrt{1-(\frac{1}{\sqrt{17} } )^{2}}$ + $\frac{1}{\sqrt{17} } \sqrt{1-(\frac{1}{{7} } )^{2}}$ ] )

= sin( $sin^{-1}$ [ $\frac{1}{7} \sqrt{1-\frac{1}{17} }$ + $\frac{1}{\sqrt{17} } \sqrt{1-\frac{1}{49} }$ ] )

= sin( $sin^{-1}$ [ $\frac{1}{7} \sqrt{\frac{16}{17} }$ + $\frac{1}{\sqrt{17} } \sqrt{\frac{48}{49} }}$ ] )

= sin( $sin^{-1}$ [ $\frac{1}{7} (\frac{4}{\sqrt{17} } )$ + $\frac{1}{\sqrt{17} } (\frac{4\sqrt{3} }{7} )$ ] )

= sin( $sin^{-1}$ [ $\frac{4 + 4\sqrt{3} }{7\sqrt{17} }$ ] )

= $\frac{4( 1 + \sqrt{3}) }{7\sqrt{17} }$

Therefore, the value of sin( $sin^{-1}$ ( $\frac{1}{7}$ ) + $tan^{-1}$ ( $\frac{1}{4}$ ) ) = $\frac{4( 1 + \sqrt{3}) }{7\sqrt{17} }$ .

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