Question

If cosec tita+cot tita = k, then prove that cos tita= k^2-1/k^2+1.

NO SMAP❌​

Answer 1

Answer:

$\mathsf{Given}$

$\mathsf{cosecθ \: + cotθ \: = k}$

$\mathsf{\frac{1}{sinθ} \: + \: \frac{cosθ}{sinθ} \: = \: k}$

$\mathsf{\frac{1 \: + cosθ}{sinθ} \: = \: k}$

$\mathsf{1 \: + \: cosθ \: = \: k sinθ}$

$\mathsf\red{Squaring \: on \: both \: sides}$

$\mathsf{{1+cosθ}^{2} \: = \: {k}^{2} {sin}^{2}θ}$

$\mathsf{{1+cosθ}^{2} \: = \: {k}^{2}{1-cosθ}^{2}}$

$\mathsf{(1+cosθ) \times (1+cosθ) \: = \: {k}^{2}.(1+cosθ) \times (1-cosθ)}$

$\mathsf{1+cosθ \: = \: {k}^{2}.(1-cosθ)}$

$\mathsf{1+cosθ \: = \: {k}^{2} - {k}^{2}cosθ}$

$\mathsf{1+(1+{k}^{2})cosθ \: = \: {k}^{2}}$

$\mathsf{(1+{k}^{2})cosθ \: = \: {k}^{2} -1}$

$\mathsf{cosθ \: = \: \frac{{k}^{2} -1}{{k}^{2} +1}}$

$\mathsf\red{ \therefore cosθ \: = \: \frac{{k}^{2} -1}{{k}^{2} +1}}$

$\mathsf\red{Hence \: Proved}$

Answer 2

Given :-

$\sf cosec \theta + cot \theta = k$

To Prove :-

$\sf cos \theta = \dfrac{k^2-1}{k^2+1}$

Solution :-

$\sf R.H.S = \dfrac{k^2-1}{k^2+1}$

       $= \sf \dfrac{( cosec \theta + cot \theta)^2-1}{(cosec \theta + cot \theta )^2+1} \\\\= \dfrac{cosec^2 \theta +cot^2 + 2 cosec \theta cot \theta -1}{cosec^2 \theta + cot^2 \theta +2cosec \theta cot \theta +1 }\\\\= \dfrac{(cosec^2 \theta -1 )+cot^2 \theta + 2cosec \theta cot \theta }{(cot^2 +1 )+cosec^2 \theta + 2cosec \theta cot \theta }$

$\bullet \bf \ cosec^2 \theta - 1 = cot^2 \theta \\\\\bullet \ cot^2 \theta + 1 = cosec^2 \theta$

        $= \sf \dfrac{cot^2 \theta+cot^2 \theta + 2cosec \theta cot \theta }{cosec^2 \theta + cosec^2 \theta + 2 cosec \theta cot \theta}\\\\= \dfrac{2cot^2 \theta + 2 cosec \theta cot \theta }{2 cosec^2 \theta+ 2 cosec \theta cot \theta }\\\\= \dfrac{2( cot^2 \theta + cosec \theta cot \theta) }{2(cosec^2 \theta +cosec \theta cot \theta )} \\\\= \dfrac{cot^2 \theta + cosec \theta cot \theta}{cosec ^2 \theta + cosec\theta cot \theta }$

$\bullet \bf \ cot \theta = \dfrac{cos \theta }{sin \theta } \\\\\bullet \ cosec \theta = \dfrac{1}{sin \theta }$

        $= \sf \dfrac{\dfrac{cos^2 \theta }{sin^2 \theta }+\left( \dfrac{1}{sin \theta } \times \dfrac{cos\theta}{sin \theta } \right)}{\dfrac{1}{sin^2 \theta } +\left( \dfrac{1}{sin \theta } \times \dfrac{cos\theta } {sin \theta } \right)} \\\\= \dfrac{\dfrac{cos^2 \theta }{sin^2 \theta } + \dfrac{cos\theta }{sin^2 \theta }}{\dfrac{1 }{sin^2 \theta }+\dfrac{cos\theta }{sin^2 \theta }}$

        $= \sf \dfrac{cos^2 \theta + cos \theta }{sin^2 \theta } \times \dfrac{sin^2 \theta }{1+cos\theta} \\\\= \dfrac{cos^2 \theta + cos \theta }{1+ cos \theta} \\\\= \dfrac{cos\theta (1+cos \theta)}{1+cos \theta} \\\\= cos \theta \\\\= R.H.S$

         

∴ L.H.S = R.H.S

Hence proved.