Question

If the angles of ∆ABC are in the ratio 1:1:2 respectively then value of secA/CosecA - Tan A / COT B​

Answer 1

I hope you have got the ans

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

in triangle ABC, we have

$\purple{\rm :\longmapsto\:\angle A : \angle B : \angle C \: = \: 1 : 1 : 2}$

Let assume that

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{\angle A = x} \\ \\ &\sf{\angle B = x}\\ \\ &\sf{\angle C = 2x} \end{cases}\end{gathered}\end{gathered}$

We know,

➢ Sum of all interior angles of a triangle is supplementary

$\rm :\longmapsto\:\angle A + \angle B + \angle C = 180\degree$

$\rm :\longmapsto\:x + x + 2x = 180\degree$

$\rm :\longmapsto\:4x = 180\degree$

$\rm\implies \:x = 45\degree$

So, we have

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{\angle A = 45\degree } \\ \\ &\sf{\angle B = x = 45\degree }\\ \\ &\sf{\angle C = 90\degree } \end{cases}\end{gathered}\end{gathered}$

Now, Consider

$\rm :\longmapsto\:\dfrac{secA}{cosecA} - \dfrac{tanA}{cotB}$

$\rm :\longmapsto\:\dfrac{sec45\degree }{cosec45\degree } - \dfrac{tan45\degree }{cot45\degree }$

$\rm \:  =  \: \dfrac{ \sqrt{2} }{ \sqrt{2} } - \dfrac{1}{1}$

$\rm \:  =  \: 1 - 1$

$\rm \:  =  \: 0$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$