Question

prove that 1+cos theta / sin theta -sin theta /1+cos theta = 2 cot theta

Answer 1

Answer:

use the sin theta = cos theta2

Answer 2

Step-by-step explanation:

$\huge\underline{\overline{\mid{\bold{\blue{\mathcal{Question:-}}\mid}}}}$

Prove that

$\huge \frac{1 + \cos(x) }{ \sin(x) } - \frac{ \sin(x) }{1 + \cos(x) } = 2 \cot(x)$

$\huge\underline{\overline{\mid{\bold{\blue{\mathcal{Hint:-}}\mid}}}}$

• We will make the LHS equal to the RHS by taking the LCM of LHS and by using few Trigonometric identities .

$\underline{\overline{\mid{\bold{\blue{\mathcal{Trigonometric \ identities:-}}\mid}}}}$

• sin^2A + cos^2A = 1

• sec^2A - tan^2A = 1

• cosec^2A = 1 + cot^2A

$\huge\underline{\overline{\mid{\bold{\blue{\mathcal{Required \ Solution:-}}\mid}}}}$

$\rightarrow\small\bf LHS:-$

$\huge \frac{1 + \cos(x) }{ \sin(x) } - \frac{ \sin(x) }{1 + \cos(x) }$

• Taking the LCM we have;

$\huge = \frac{(1 + { cos \: x )}^{2} - { \sin}^{2}x }{ sin \: x(1 + cos \: x) }$

• Using the identity of (a+b)^2 = a^2 + b^2 + 2ab we have;

$\huge \frac{1 + {cos}^{2} x + 2cos \: x - {sin}^{2}x }{sin \: x(1 + cos \: x)}$

$\mathcal{\green{Now,}}$

we know that

$1 - {sin}^{2} x = {cos}^{2} x$

• Using this identity we have;

$\huge \frac{ {cos}^{2} x + 2cos \: x + (1 - sin {}^{2}x) }{sin \: x(1 + cos \: x) }$

$\huge \frac{ {cos}^{2} x + 2cos \: x + {cos}^{2}x }{sin \: x(1 + cos \: x)}$

$\huge = \frac{2 {cos}^{2} x + 2cos \: x}{sin \: x(1 + cos \: x)}$

$\huge = \frac{2 \: cos \: x(cos \: x + 1)}{sin \: x(1 + cos \: x)}$

$\huge = \frac{2 {cos} \: x(1 + cos \: x) }{sin \: x(1 + cos \: x)}$

$\huge = \frac{2 \: cos \: x}{sin \: x} = 2cot \: x$

• because cos x / sin x = cot x.

$\rightarrow\small\bf RHS$

□ Hence, proved.

$\mathfrak{\red{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ @MissTranquil}}$

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☆ Note:-

• Here, theta is taken as 'x'.