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(cosecA-sinA)(secA-cosA) =1/tanA+cotA

(cosecA-sinA)(secA-cosA) =1/tanA+cotA

Grade:10

4 Answers

Nishant Vora IIT Patna
askIITians Faculty 2467 Points
6 years ago
Hello Student, Please find the solution

(cosecA-sinA)(secA-cosA)
=(1/sinA – sinA )(1/cosA – cosA)
=[(1- sin2A)/sinA][(1-cos2A)/cosA]
=[(cos2A)/sinA][(sin2A)/cosA]
=sinA*cosA
\frac{sinA cosA}{sin^2A + cos^2A}

= \frac{1}{sin^2A/sinA cosA + cos^2A/sinA cosA }

=\frac{1}{tanA + cotA }
=rhs

I hope it is clear
Thanks

Animesh singh
35 Points
3 years ago
LHS:-(cosecA-sinA) (secA-cosA)(1/sinA-sinA)(1/cosA-cosA)(1-sin²A/sinA) (1-cos²A/cosa)(Cos²A/sinA)(sin²A/cosA)SinA.cosaRHS:-1/tanA+cotA1/(sinA/cosa+cosa/sinA)1/(sin²A+cos²A/sinAcosA)SinA.cosA/1SinA.cosa LHS=RHShence proved.... Answered by:- Animesh singh (MASTERMIND)
Animesh singh
35 Points
3 years ago
LHS:-(cosecA-sinA) (secA-cosA)=(1/sinA-sinA)(1/cosA-cosA)=(1-sin²A/sinA) (1-cos²A/cosa)=(Cos²A/sinA)(sin²A/cosA)=SinA.cosaRHS:-1/tanA+cotA=1/(sinA/cosa+cosa/sinA)=1/(sin²A+cos²A/sinAcosA)=SinA.cosA/1=SinA.cosa LHS=RHShence proved.... Answered by:- Animesh singh (MASTERMIND)
Komal yadav
15 Points
2 years ago
Taking LHS: (1/sinA-sinA/1) (1/cosA-cosA/1) 
=(1-sin2A/sinA) (1-cos2A/cosA)
=(cos2A/sinA) (sin2A/cosA)
=(cos2A.sin2A/sinA.cosA)          
=Taking RHS: 1/sinA/cosA+cosA/sinA
=Taking LCM sin2A+cos2A/sinA.cosA
sinA. cosA
Hence proved

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