cos20/(1-tan0) + sin30 /(sin 0 - cos0) = (1+sin0 cos 0) - Maths - Some Applications of Trigonometry

Question

cos2 0/(1-tan0) +
sin3 0 /(sin 0 -
cos0) = (1+sin0 cos
0)

Ajanta Trivedi · Accepted Answer

LHS of the given equation is:

$\frac{\cos^{2} θ}{1 - \tan θ} + \frac{\sin^{3} θ}{\sin θ - \cos θ} = \frac{\cos^{2} θ}{1 - \frac{\sin θ}{\cos θ}} + \frac{\sin^{3} θ}{\sin θ - \cos θ} = \frac{\cos^{2} θ}{(\cos θ - \sin θ) / \cos θ} + \frac{\sin^{3} θ}{\sin θ - \cos θ} = \frac{\cos^{3} θ}{\cos θ - \sin θ} - \frac{\sin^{3} θ}{\cos θ - \sin θ} = \frac{\cos^{3} θ - \sin^{3} θ}{(\cos θ - \sin θ)} = \frac{(\cos θ - \sin θ) (\cos^{2} θ + \sin^{2} θ + \cos θ \sin θ)}{\cos θ - \sin θ} [∵ a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b)]$

$= \cos^{2} θ + \sin^{2} θ + \cos θ \sin θ = 1 + \sin θ \cos θ [∵ \sin^{2} θ + \cos^{2} θ = 1]$

=RHS

hope this helps you.

cheers!!

cos2 0/(1-tan0) + sin3 0 /(sin 0 - cos0) = (1+sin0 cos 0)

cos² 0/(1-tan0) + sin³ 0 /(sin 0 - cos0) = (1+sin0 cos 0)