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# Inverse circular functions,Principal values of sin−1x,cos−1x,tan−1x.tan−1x+tan−1y=tan−1x+y1−xy, xy<1 π+tan−1x+y1−xy, xy>1.(a) sin−11213+cos−145+tan−16316=π(b) 2cos−13√13+cot−11663+12cos−1725=π

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Solution

## (a) Since sin−11213=tan−1125and cos−145=tan−134, we haveL.H.S. =tan−1125+tan−134+tan−16316and since 125×34>1, we have tan−1125+tan−134=π+tan−1(12/5)+(3/4)1−(12/5).(3/4)=π+tan−1{−6316}=π+tan−16316 [∵tan−1(−x)=−tan−1x]HenceL.H.S. =π−tan−16316+tan−16316=π= R.H.S.(b) As in part (a), we will change each term to tan−1 2cos−13√13=cos−1(2x2−1) =cos−1513=tan−11213 cot−11663=tan−16316 12cos−1725=θsay, ∴ 2θ=725or 1−t21+t2=725 ∴t2=916 or t=34,−34Since cos2θ=+ive 0≤2θ<π∴ 0≤θ<π/2and hence t=tanθ=+ive ∴tanθ=3/4∴ L.H.S. =tan−11213+tan−16316+tan−134Rest as in part (a).

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