Question

if sin theta + cos theta is equal to a , tan theta + cot theta is equal to b show that a square minus 1 by 2 is equal to 1 by b​

Answer 1

Answer:

Let theta = A

So LHS :

=((sinA + cosA)^2 -1)/2

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

= (1 + 2 sinAcosA -1)/2

= sinAcosA

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

[As (sin^2 A + cos^2 A)=1 ]

[Now dividing numerator and denominator by (sinAcosA) ]

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

=1/(TanA + CotA)

= 1/b = RHS

Hence Proved

PLS MARK BRAINLIEST AND FOLLOW ME FOR MORE ANSWERS.

Answer 2

Answer:

1/b

please

mark as brilliant