Question

prove that sin thetha - cos thetha/sin thetha +sin thetha + cos thetha / sin thetha - cos thetha = 2/2 sin square thetha -1 ​

Answer 1

Answer:

$\huge\mathcal{\green{Hola!}}$

Step-by-step explanation:

☆ To prove:-

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

☆ Trigonometric identities:-

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

${sec}^{2} x - {tan}^{2} x = 1$

${cosec}^{2} x - 1 = {cot}^{2} x$

☆ Required Solution:-

･➝ L.H.S

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

･➝ Taking their LCM we have;

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

[ because (a+b)(a-b) = a^2 - b^2 ]

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

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

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

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

$\huge\mathcal{\green{Therefore,}}$

$\frac{1 - 2sin \: x.cos \: x + \: 1 + \: 2sinx.cos \: x}{ {sin}^{2}x - {cos}^{2} x } = \frac{2}{ {sin}^{2}x - {cos}^{2} x }$

$\huge\mathcal{\green{Also}}$

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

$\huge\mathcal{\green{Therefore,}}$

$= > \frac{2}{ {sin}^{2} x \: - (1 - {sin}^{2}x) }$

$\huge= > \frac{2}{ {sin}^{2} x + {sin}^{2} x - 1 } = \frac{2}{2 \: {sin}^{2} x - 1}$

= R.H.S

□ Hence, proved.

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$\huge\fbox{\orange{ be \ brainly}}$

Answer 2

$\huge{\mathcal{\purple{A}\green{N}\pink{S}\blue{W}\purple{E}\green{R}\pink{!}}}!$

above is the right answer