Question

( cosecA -sin A ) ( sec A - cos A ) = 1/ tan A + cot A​

Answer 1

Solution is in attachment ..........

Answer 2

Step-by-step explanation:

L.H.S

= ( cosec A - sin A ) (sec A - cos A )

But, we know that,

• cosec A = 1/sin A
• sec A = 1/cos A

Therefore, we will get,

= (1/sinA - sinA)(1/cosA - cosA)

= (1- sin^2A)/sinA (1-cos^2A)/cosA

= (1-sin^2A)(1-cos^2A)/(sinAcosA)

= (1 - cos^2A - sin^2A + cos^2A sin^2A)/(sinAcosA)

= {1-(sin^2A+cos^2A)+ sin^2A cos^2A}/(sinAcosA)

= (1-1+sin^2Acos^2A)/(sinAcosA)

= (sin^2A cos^2A)/(sinA cosA)

= sinA cosA

= sinA cosA/1

= (sinA cosA) /(sin^2 A + cos^2 A)

= 1/(sin^2A+cos^2A)/(sinA cosA)

= 1/{(sin^2A/sinAcosA)+(cos^2A/sinAcosA)}

= 1/{(sinA/cosA)+(cosA/sinA)}

= 1/(tanA + cotA)

= R.H.S

Thus, L.H.S = R.H.S

Hence, Proved.