Question

prove that 1+ cos teata / sin teata - sin teata / 1+ cos teata = 2 cot teata​

Answer 1

$\bold{ \large\sf \boxed{ \boxed{ To \: \: Prove}}}$

$\dfrac{1 + cos \: \theta}{sin \: \theta} - \dfrac{sin \theta }{1 + cos \theta } = 2cot \theta$

LHS :-)

$\sf\dfrac{1 + cos \: \theta} {sin \: \theta} - \dfrac{sin \theta }{1 + cos \theta } \\ \\ = \sf \dfrac{(1 + cos \theta) {}^{2} - {sin}^{2} \theta}{sin \theta \: (1 + cos \theta) } \\ \\ = \sf \frac{1 + {cos}^{2} \theta + 2cos \theta - {sin}^{2} \theta }{sin \theta \: (1 + cos \theta)} \\ \\ = \sf \frac{1 + {cos}^{2} \theta + 2cos \theta - 1 + {cos}^{2} \theta }{sin \theta \: (1 + cos \theta)} \\ \\ = \sf \dfrac{2cos \theta + 2 {cos}^{2} \theta }{sin \theta(1 + cos \theta) } \\ \\ = \sf \frac{2cos \theta (1 + cos \theta) }{sin \theta(1 + cos \theta)} \\ \\ = \sf2cot \theta$

=RHS

Answer 2

Answer:

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