Question

1+cos \sin-sin/1+cos=2cot​

Answer 1

Answer:

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$\frac{1 + \cos }{ \sin } - \frac{ \sin }{1 + \cos } = 2 \cot \\ \frac{ {(1 + \cos) }^{2} - { \sin }^{2} }{ \sin(1 + \cos) } = 2 \cot \\ \\ \frac{1 + { \cos }^{2} + 2 \cos - { \sin }^{2} }{ \sin(1 + \cos) } = 2 \cot \\ \frac{(1 - { \sin}^{2}) + { \ { \cos }^{2} + 2 \cos } }{ \sin(1 + \cos) } = 2 \cot \\ \frac{ { \cos }^{2} + { \cos }^{2} + 2 \cos }{ \sin(1 + \cos)} = 2 \cot \\ \frac{2 { \cos}^{2} + 2 \cos }{ \sin(1 + \cos)} = 2 \cot \\ \frac{2 \cos(1 + \cos) }{ \sin(1 + \cos)} = 2 \cot \\ \frac{2 \cos }{ \sin } = 2 \cot \\ 2 \times \frac{ \cos }{ \sin } = 2cot \\ 2cot = 2cot$

L. H. S = R. H. S

Hence proved.

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Answer 2

$\huge{\underline{\underline{\bf{\red{Solution}}}}}$

To prove : -

$\implies\sf \frac{1+cos\theta}{sin\theta}-\frac{sin\theta}{1+cos\theta}=2cot$

Proof : -

LHS

$\implies\sf \large\frac{(1+cos\theta)^2-sin^2\theta}{sin\theta(1+cos\theta)}$

$\implies\sf \frac{(1+cos^2\theta+2cos\theta)-sin^2\theta}{sin\theta(1+cos\theta)}$

$\implies\sf \frac{(1-sin^2\theta)+cos^2+2cos\theta}{sin\theta(1+cos\theta)}$

$\implies\sf \frac{cos^2\theta+cos^2\theta+2cos\theta}{sin\theta(1+cos\theta)}$

$\implies\sf \frac{2cos\theta+2cos^2\theta}{sin\theta(1+cos\theta)}$

$\implies\sf \frac{2cos\theta(1+cos\theta}{sin\theta(1+cos\theta)}$

$\implies\sf \frac{2cos\theta}{sin\theta}$

$\large\implies\sf 2cot$ =RHS proved

Note:-

• cos∅/sin∅ = cot∅
• (a+b)² = a²+b²+2ab
• sin²∅+cos²∅ = 1