(sin^2A+cosec^2A)^2+(cosA+secA)^2-tan^2A-cot^2A
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(sin^2 A + 1/sin ^2 A )^2+ (cosA + 1/cosA)^2 - (tan^2 A + 1/tan^2 A)
(sin A + cosec A)2 + (cos A+ sec A)2 -------->
eq(I)
= (sin2 A + 2 sin A . cosec A + cosec2 A) + (cos2 A + 2 cos A . sec A + sec2 A)
= sin2 A + 2 + cosec2 A + cos2 A + 2 + sec2 A ---(sin A . cosec A =1=cos A . sec A)
= sin2 A + 2 + 1 + cot2 A + cos2 A + 2 + 1 + tan2 A ---(cosec2 A=1+ cot2 A & sec2A=1+tan2 A)
= ( sin2 A + cos2 A ) + 6 + ( cot2 A + tan2 A )
= 1 + 6 + cot2 A + tan2 A-----(sin2 A + cos2 A =1)
= 7 + tan2 A + cot2 A
= 7 + tan2 A + cot2 A- tan2 A - cot2 A
= 7
(sin A + cosec A)2 + (cos A+ sec A)2 -------->
eq(I)
= (sin2 A + 2 sin A . cosec A + cosec2 A) + (cos2 A + 2 cos A . sec A + sec2 A)
= sin2 A + 2 + cosec2 A + cos2 A + 2 + sec2 A ---(sin A . cosec A =1=cos A . sec A)
= sin2 A + 2 + 1 + cot2 A + cos2 A + 2 + 1 + tan2 A ---(cosec2 A=1+ cot2 A & sec2A=1+tan2 A)
= ( sin2 A + cos2 A ) + 6 + ( cot2 A + tan2 A )
= 1 + 6 + cot2 A + tan2 A-----(sin2 A + cos2 A =1)
= 7 + tan2 A + cot2 A
= 7 + tan2 A + cot2 A- tan2 A - cot2 A
= 7
Ishita333:
Not in integer value but it cn b simplified in single term
Take Z=theta. If, tanZ=A/B, then find, sinZ-BcosZ)/(AsinZ+BsinZ), Z belongs to 1st quadrant.
Ye bhi solve krdo..plzx
Mera ans..cos2theta aa rha h
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