Question

1) sin Acos A(tan A + cotA) = 1​

Answer 1

Answer:

Step-by-step explanation:

LHS

= $sinAcosA(tanA + cotA)$

= $sinAcosA(\frac{sinA}{cosA} + \frac{cosA}{sinA}  )$

= $sinAcosA(\frac{sin^{2}A  + cos^{2} A}{sinAcosA} )$

Since $sin^{2}A  + cos^{2}A  = 1$

LHS = $sinAcosA(\frac{1}{sinAcosA})$

LHS = 1

LHS = RHS

Hence, proved!

Answer 2

Step-by-step explanation:

=sinAcosA(sinA/cosA+cosA/sinA)

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

=sin^2A+cos^2A

=1