Question

sin theta/sec theta + 1 + sin theta / sec theta -1= 2cot theta​

Answer 1

TO PROVE :-

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

SOLUTION :-

We have ,

$\\ \sf \: L.H.S = \dfrac{sin}{sec + 1} + \dfrac{sin}{sec - 1}$

Cross multiplying,

$\\ \implies \sf \: \dfrac{sin(sec - 1) + sin(sec + 1)}{(sec + 1)(sec - 1)}$

In denominator,

★ (a+b)(a-b) = a² - b²

• a = secø
• b = 1

Putting values,

$\\ \implies \sf \: \dfrac{sin\theta.sec\theta - \cancel{sin\theta}+ sin\theta.sec\theta + \cancel{sin\theta}}{ {sec}^{2}\theta - {1}^{2} } \\ \\ \implies \sf \: \dfrac{2sin\theta.sec\theta}{ {sec}^{2}\theta - 1 }$

$\\ \bigstar \boxed{ \bf \: 1 + {tan}^{2}\theta = {sec}^{2} \theta } \\ \\ \therefore \bf \: \underline{{sec}^{2}\theta - 1 = {tan}^{2} \theta}$

Using above formula ,

$\\ \implies \sf \: \dfrac{2sin\theta.sec\theta}{ {tan}^{2}\theta }$

$\\ \bigstar \boxed{ \bf \:sec\theta = \dfrac{1}{cos\theta} } \: \: \: \: \:$

Using above formula,

$\\ \implies \sf \: \dfrac{2 \times sin\theta \times \dfrac{1}{cos\theta} }{ {tan}^{2}\theta }$

$\\ \bigstar \boxed{ \bf \: \dfrac{sin\theta}{cos\theta} = tan\theta}$

Using above formula ,

$\\ \implies \sf \: \dfrac{2tan\theta}{ { {tan}^{2}\theta } } \\ \\ \implies \sf \: \dfrac{2}{tan\theta}$

$\\ \bigstar\boxed{ \bf \: tan\theta = \dfrac{1}{cot\theta} }$

$\\ \implies \sf \: 2cot\theta \: \: = R.H.S \: \: \: \: \: \: \: \: \: (verified)$

MORE FORMULAS :-

$\\ \dag \tt \: \: \: {sin}^{2}\theta + {cos}^{2}\theta = 1 \\ \\ \dag \tt \: \: \: 1 + {tan}^{2}\theta = {cosec}^{2}\theta \\ \\ \dag \tt \: \: \: sin2\theta = 2sin\theta.cos\theta \\ \\ \dag \tt \: \: \: cos2\theta = {cos}^{2}\theta - {sin}^{2}\theta$