Question

If cot theta/2 =2.5 then find cosec theta.

Answer 1

GIVEN :–

$\\ \implies \bf \cot \left(\dfrac{ \theta}{2}\right) = 2.5$

TO FIND :–

$\\ \implies \bf cosec(\theta) = \: ?$

SOLUTION :–

$\\ \implies \bf \cot \left(\dfrac{ \theta}{2}\right) = 2.5$

$\\ \implies \bf \tan \left(\dfrac{ \theta}{2}\right) = \dfrac{1}{2.5 }$

$\\ \implies \bf \tan \left(\dfrac{ \theta}{2}\right) = \dfrac{2}{5}$

• We know that –

$\\ \implies \pink{ \boxed{ \bf \tan( \theta)= \dfrac{2\tan \left(\dfrac{ \theta}{2}\right)}{1 - \tan^{2} \left(\dfrac{ \theta}{2}\right)}}}$

• Put the values –

$\\ \implies\bf \tan( \theta)= \dfrac{2\left(\dfrac{2}{5} \right)}{1 - \left(\dfrac{2}{5} \right)^{2}}$

$\\ \implies\bf \tan( \theta)= \dfrac{\dfrac{4}{5}}{1 -\dfrac{4}{25}}$

$\\ \implies\bf \tan( \theta)= \dfrac{\dfrac{4}{5}}{\dfrac{25 - 4}{25}}$

$\\ \implies\bf \tan( \theta)= \dfrac{\dfrac{4}{5}}{\dfrac{21}{25}}$

$\\ \implies\bf \tan( \theta)= \dfrac{4}{5} \times \dfrac{25}{21}$

$\\ \implies\bf \tan( \theta)= 4\times \dfrac{5}{21}$

$\\ \implies\bf \tan( \theta)= \dfrac{20}{21}$

• We should write this as –

$\\ \implies\bf \dfrac{ \sin\theta}{ \cos\theta}= \dfrac{20}{21}$

$\\ \implies\bf \dfrac{ \sin\theta}{ \sqrt{1 -\sin^{2} \theta}}= \dfrac{20}{21}$

$\\ \implies\bf \dfrac{ \sin^{2} \theta}{ {1 -\sin^{2} \theta}}= \dfrac{400}{441}$

$\\ \implies\bf441\sin^{2} \theta = 400(1 -\sin^{2} \theta)$

$\\ \implies\bf441\sin^{2} \theta = 400 -400\sin^{2} \theta$

$\\ \implies\bf441\sin^{2} \theta + 400\sin^{2} \theta= 400$

$\\ \implies\bf841\sin^{2} \theta= 400$

$\\ \implies\bf\sin^{2} \theta= \dfrac{400}{841}$

$\\ \implies\bf\sin\theta= \pm\sqrt \dfrac{400}{841}$

$\\ \implies\bf\sin\theta=\pm \dfrac{20}{29}$

$\\ \implies\bf \dfrac{1}{\sin\theta}=\pm \dfrac{29}{20}$

$\\ \implies \large{ \boxed{\bf cosec(\theta)= \pm \dfrac{29}{20}}}$

Answer 2

GIVEN :–

$\\ \implies \bf \cot \left(\dfrac{ \theta}{2}\right) = 2.5$

TO FIND :–

$\\ \implies \bf cosec(\theta) = \: ?$

SOLUTION :–

$\\ \implies \bf \cot \left(\dfrac{ \theta}{2}\right) = 2.5$

$\\ \implies \bf \tan \left(\dfrac{ \theta}{2}\right) = \dfrac{1}{2.5 }$

$\\ \implies \bf \tan \left(\dfrac{ \theta}{2}\right) = \dfrac{2}{5}$

• We know that –

$\\ \implies \pink{ \boxed{ \bf \tan( \theta)= \dfrac{2\tan \left(\dfrac{ \theta}{2}\right)}{1 - \tan^{2} \left(\dfrac{ \theta}{2}\right)}}}$

• Put the values –

$\\ \implies\bf \tan( \theta)= \dfrac{2\left(\dfrac{2}{5} \right)}{1 - \left(\dfrac{2}{5} \right)^{2}}$

$\\ \implies\bf \tan( \theta)= \dfrac{\dfrac{4}{5}}{1 -\dfrac{4}{25}}$

$\\ \implies\bf \tan( \theta)= \dfrac{\dfrac{4}{5}}{\dfrac{25 - 4}{25}}$

$\\ \implies\bf \tan( \theta)= \dfrac{\dfrac{4}{5}}{\dfrac{21}{25}}$

$\\ \implies\bf \tan( \theta)= \dfrac{4}{5} \times \dfrac{25}{21}$

$\\ \implies\bf \tan( \theta)= 4\times \dfrac{5}{21}$

$\\ \implies\bf \tan( \theta)= \dfrac{20}{21}$

• We should write this as –

$\\ \implies\bf \dfrac{ \sin\theta}{ \cos\theta}= \dfrac{20}{21}$

$\\ \implies\bf \dfrac{ \sin\theta}{ \sqrt{1 -\sin^{2} \theta}}= \dfrac{20}{21}$

$\\ \implies\bf \dfrac{ \sin^{2} \theta}{ {1 -\sin^{2} \theta}}= \dfrac{400}{441}$

$\\ \implies\bf441\sin^{2} \theta = 400(1 -\sin^{2} \theta)$

$\\ \implies\bf441\sin^{2} \theta = 400 -400\sin^{2} \theta$

$\\ \implies\bf441\sin^{2} \theta + 400\sin^{2} \theta= 400$

$\\ \implies\bf841\sin^{2} \theta= 400$

$\\ \implies\bf\sin^{2} \theta= \dfrac{400}{841}$

$\\ \implies\bf\sin\theta= \pm\sqrt \dfrac{400}{841}$

$\\ \implies\bf\sin\theta=\pm \dfrac{20}{29}$

$\\ \implies\bf \dfrac{1}{\sin\theta}=\pm \dfrac{29}{20}$

$\\ \implies \large{ \boxed{\bf cosec(\theta)= \pm \dfrac{29}{20}}}$