Question

Guys plz help me out of this question. Explain with full solution step-by-step-explanation ​

Answer 1

Topic :-

Trigonometry

Given :-

$\sf {6sec\alpha -5tan\alpha =k}$

$\sf {6sec\alpha +ktan\alpha =5}$

To Find :-

$\sf{k^2}$

Solution :-

Subtracting the equations,

$\sf {6sec\alpha +ktan\alpha - (6sec\alpha -5tan\alpha) =5-k}$

$\sf {6sec\alpha +ktan\alpha - 6sec\alpha +5tan\alpha=5-k}$

$\sf {ktan\alpha +5tan\alpha =5-k}$

$\sf {tan\alpha(5+k) =5-k}$

$\sf {tan\alpha=\dfrac{5-k}{5+k}}$

$\sf{sec\alpha= \sqrt{1+tan^2\alpha}}$

$\sf {sec\alpha= \sqrt{1+\left(\dfrac{5-k}{5+k}\right)^2}}$

$\sf {sec\alpha= \sqrt{\left(\dfrac{(5+k)^2+(5-k)^2}{(5+k)^2}\right)}}$

$\sf {sec\alpha= \sqrt{\left(\dfrac{(50+2k^2)}{(5+k)^2}\right)}}$

Put the obtained values in any equation :-

$\sf {6sec\alpha -5tan\alpha =k}$

$\sf {6{\left(\dfrac{ \sqrt{50+2k^2}}{(5+k)}\right)}-5\left(\dfrac{5-k}{5+k}\right ) =k}$

$\sf {\dfrac{6 {\sqrt{50+2k^2}-5(5-k)}}{5+k}=k}$

$\sf {6 \sqrt{50+2k^2}-25+5k=k(5+k)}$

$\sf {6 \sqrt{50+2k^2}-25+5k=5k+k^2}$

$\sf {6 \sqrt{50+2k^2}=k^2+25}$

Square it,

$\sf {({6 \sqrt{50+2k^2})}^2={(k^2+25)}^2}$

$\sf {36(50+2k^{2})=k^{4}+50k^{2}+625}$

$\sf { 1800+72k^{2}=k^{4}+50k^{2}+625}$

$\sf{ k^{4}-22k^{2}-1175=0}$

Solve it using Quadratic Formula,

$\sf { k^2=\dfrac{22 \pm \sqrt{(22)^{2}-4(-1175)}}{2}}$

$\sf { k^2=\dfrac{22 \pm \sqrt{5184}}{2}}$

$\sf { k^2=\dfrac{22 \pm 72}{2}}$

$\sf { k^2={11 \pm 36}}$

$\sf {k^2=47\:or\:-25}$

We will reject the negative value as it can't be negative.

Answer :-

So, value of $\sf{k^2}$ is 47.

Answer 2

Answer:

47

Step-by-step explanation:

note : please kindly refer to above attachment