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

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

rachel , 10 Years ago

Grade 10

4 Answers

Nishant Vora

Last Activity: 10 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

Last Activity: 7 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

Last Activity: 7 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)