Question

prove that ,1+cos 2a=cot a sin2 a

Answer 1

from property.
Cos 2A = 2cos^2A-1
sin2a = 2sinacosa

evaluate the eqn LHS=rhs

Answer 2

Answer:

$\bold{\underline{To \ Proof:}}$

$\sf \dfrac{1 + cos \theta - {sin}^{2} \theta}{sin \theta(1 + cos \theta)} = cot \theta$

$\bold{\underline{Proof:}}$

LHS:

$\sf \implies \dfrac{1 + cos \theta - {sin}^{2} \theta}{sin \theta(1 + cos \theta)} \\ \\ \sf {sin}^{2} \theta = 1 - {cos}^{2} \theta : \\ \sf \implies \dfrac{1 + cos \theta - (1 - {cos}^{2} \theta )}{sin \theta(1 + cos \theta)} \\ \\ \sf (1 - cos ^{2} \theta ) = ( {1}^{2} - cos ^{2} \theta ) = (1 - cos \theta )(1 + cos \theta ) : \\ \sf \implies \dfrac{1 + cos \theta - (1 - cos \theta )(1 + cos \theta )}{sin \theta(1 + cos \theta)} \\ \\ \sf \implies \dfrac{ \cancel{1 + cos \theta}(1 - (1 - cos \theta ))}{sin \theta( \cancel{1 + cos \theta})} \\ \\ \sf \implies \dfrac{1 - (1 - cos \theta )}{sin \theta} \\ \\ \sf \implies \dfrac{1 - 1 + cos \theta )}{sin \theta} \\ \\ \sf \implies \dfrac{cos \theta }{sin \theta} \\ \\ \sf \implies cot \theta$

RHS:

$\sf \implies cot \theta$

$\therefore$

$\bold{LHS = RHS}$

$\underline{ \sf Hence \: Proved}$