Question

pls answer this question
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Answer 1

Answer:

Given :

$\bullet \ \mathrm{sinx + siny = \dfrac{1}{4}}$

$\bullet \ \mathrm{cosx + cosy = \dfrac{1}{3}}$

To Prove :

$\bullet \ \mathrm{tan\bigg(\dfrac{x + y}{2}\bigg)= \dfrac{3}{4}}$

$\bullet \ \mathrm{cot(x+y) = \dfrac{7}{24}}$

Solution :

We know that,

$\begin{gathered}\boxed{\begin{aligned} \mathrm { \bullet \ sinC + sinD} = & \mathrm{2sin\bigg(\dfrac{C+D}{2}\bigg)cos\bigg(\dfrac{C-D}{2}\bigg)} \\\frac{\qquad \qquad \qquad \qquad }{} & \frac{\qquad \qquad \qquad \qquad\qquad\qquad }{} \\ \mathrm { \bullet \ cosC + cosD} = & \mathrm{2cos\bigg(\dfrac{C+D}{2}\bigg)cos\bigg(\dfrac{C-D}{2}\bigg)} \\ \end{aligned}} \end{gathered}$

As here,

$\longmapsto \mathrm{sinx + siny = \dfrac{1}{4}}\ \text{and} \ \mathrm{cosx + cosy = \dfrac{1}{3}}$

$\implies \mathrm {sinx + siny = 2sin\bigg(\dfrac{x+y}{2}\bigg)cos\bigg(\dfrac{x-y}{2}\bigg) = \dfrac{1}{4}} \ \ \ ---- [1]$

$\implies \mathrm {cosx + cosy = 2cos\bigg(\dfrac{x+y}{2}\bigg)cos\bigg(\dfrac{x-y}{2}\bigg) = \dfrac{1}{3}} \ \ \ - - - - [2]$

Dividing 1 by 2, we get,

$\implies \mathrm{\dfrac{2sin\bigg(\dfrac{x+y}{2}\bigg)cos\bigg(\dfrac{x-y}{2}\bigg)}{2cos\bigg(\dfrac{x+y}{2}\bigg)cos\bigg(\dfrac{x-y}{2}\bigg)} = \dfrac{\bigg(\dfrac{1}{4}\bigg)}{\bigg(\dfrac{1}{3}\bigg)}}$

$\implies \mathrm{\dfrac{\not2sin\bigg(\dfrac{x+y}{2}\bigg)\bcancel{cos\bigg(\dfrac{x-y}{2}\bigg)}}{\not2cos\bigg(\dfrac{x+y}{2}\bigg)\bcancel{cos\bigg(\dfrac{x-y}{2}\bigg)}} = \dfrac{3}{4}}$

$\implies \mathrm{tan\bigg(\dfrac{x+y}{2}\bigg) = \dfrac{3}{4}}$

Hence Proved!!

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Now,

$\longmapsto\mathrm{cot(x+y) = cot\bigg(\dfrac{2(x+y)}{2}\bigg)}$

Let (x+y)/2 be t

we know that,

$\boxed{\mathrm{cot\theta = \dfrac{1}{tan\theta}}}$

$\implies \mathrm{\dfrac{1}{tan2t}}$

Now, let us find tan2t then reciprocal of that will be the desired answer.

We know that,

$\boxed{\mathrm{tan\theta = \dfrac{2tan\dfrac{\theta}{2}}{1 + tan^2\dfrac{\theta}{2}}}}$

here, θ = t = (x+y)/2

$\longmapsto \mathrm{tan2t = tan\bigg(\dfrac{2(x+y)}{2}\bigg) = \dfrac{2\times \dfrac{3}{4}}{1 - \bigg(\dfrac{3}{4}\bigg)^2}}$

$\implies \mathrm{tan\bigg(\dfrac{2(x+y)}{2}\bigg) = \dfrac{6 \times 4}{16 - 9}}$

$\implies \mathrm{tan\bigg(\dfrac{2(x+y)}{2}\bigg) = \dfrac{24}{7}}$

$\implies \mathrm{tan2t = \dfrac{24}{7}}$

Recalling,

$\longmapsto \mathrm{cot(x+y) = \dfrac{1}{tan2t}}$

$\implies \mathrm{cot(x+y) = \dfrac{1}{\bigg(\dfrac{24}{7}\bigg)}}$

$\implies \mathrm{cot(x+y) = \dfrac{7}{24}}$

Hence proved!!

Hope it helps!!