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AAKASH SERIES-INVERSE TRIGONOMETRIC FUNCTIONS-EXERCISE I
- Sin^(-1)("sin"(2pi)/3)=
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- cos^(-1)("cos"(7pi)/6) is equal to
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- sin^(-1)((sqrt(3))/2)-tan^(-1)(-sqrt(3))=
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- sec^(2)(Tan^(-1)2)+cosec^(2)(Cot^(-1)3)=
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- cos[sin^(-1)(-4//5)-cos^(-1)(4//5)]=
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- sec^(-1)((2sqrt(2))/(1+sqrt(3)))=
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- I. sin("Cos"^(-1)3/5)=4/5 II. cos("Tan"^(-1)7/24)=24/25
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- cos(sin^(-1)63//65)=
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- cos^(-1)(8/17)-cos^(-1)(5/13)=
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- Sin^(-1)(4/5)-Sin^(-1)(5/13)=
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- tan^(-1)(2+)tan^(-1)(3)=
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- 2tan^(-1)(1//2)+sin^(-1)(3//5)=
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- The value of tan^(-1)(m/n)-tan^(-1)((m-n)/(m+n)) is
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- 2"tan"^(-1)1/5+"tan"^(-1)1/7+2"tan"^(-1)1/8=
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- cos[2"tan"^(-1)1/5+"cos"^(-1)63/65]=
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- tan(1/2Cos^(-1)0)=
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- sin(2tan^(-1)(8/15))=
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- cos(2cos^(-1)(7//25))=
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- If tan^(-1)x+tan^(-1)y+tan^(-1)z=pi then x+y+z=
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- If cos^(-1) x + cos^(-1) y + cos^(-1) z = 3pi , then
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