Question

$\huge \underline\mathfrak\color{lime}question$


$\frac{ \sin(a) }{ \cos(a) } + \frac{ \tan(a) }{ \cot(a) } = \frac{ \sec(a) }{ \sin(a) }$

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Answer 1

Question:-

• Solve the following problem

Answer:-

• The value of alpha ( α ) is 45°

Explanation:-

Given that:-

$\bullet \; {\underline{\boxed{\bf{ \frac{\sin \alpha }{\cos \alpha } + \frac{\tan \alpha }{\cot \alpha } = \frac{\sec \alpha }{\sin \alpha } }}}}$

To Find:-

• The value of alpha ( a )

Required Solution :-

─ Now let's re - write what we have

$\longrightarrow \sf \dfrac{\sin \alpha }{\cos \alpha } + \dfrac{\tan \alpha }{\cot \alpha } = \dfrac{\sec \alpha }{\sin \alpha }$

─ Evaluating further we get

$\longrightarrow \sf \tan \alpha + \dfrac{\tan \alpha }{\dfrac{1}{\tan \alpha } } = \dfrac{\sec \alpha }{\dfrac{1}{\sec \alpha } }$

$\longrightarrow \sf \tan \alpha + \tan \alpha \; . \; \dfrac{\tan \alpha }{1} = \sec \alpha \; . \; \dfrac{\sec \alpha }{1}$

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

$\longrightarrow \sf \tan \alpha = \sec^2 \alpha - \tan^2\alpha$

$\longrightarrow \sf \tan \alpha = 1$

$\longrightarrow \sf \tan \alpha = \tan ( 45 )$

─ Comparing both the sides and dropping the tangent function

$\longrightarrow \rm {\red{\underline{\underline{The \; value \; of \;\alpha = ( 45 ) \degree }}}}$

Henceforth:-

• The value of alpha ( α ) is equal to 45°

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Answer 2

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

Given Trigonometric equation is

$\rm :\longmapsto\:\dfrac{sina}{cosa} + \dfrac{tana}{cota} = \dfrac{seca}{sina}$

can be rewritten as using

$\boxed{\tt{ tanx = \frac{sinx}{cosx}}} \: \: \\ \\ \boxed{\tt{ cotx = \frac{1}{tanx}}} \\ \\ \boxed{\tt{ secx = \frac{1}{cosx}}}$

So, we have

$\rm :\longmapsto\:tana + {tan}^{2}a = \dfrac{1}{sina \: cosa}$

can be further rewritten as

$\rm :\longmapsto\:tana + {tan}^{2}a = \dfrac{ {sin}^{2}a + {cos}^{2}a}{sina \: cosa}$

$\rm :\longmapsto\:tana + {tan}^{2}a = \dfrac{ {sin}^{2}a}{sina \: cosa} + \dfrac{ {cos}^{2}a }{sina \: cosa}$

$\rm :\longmapsto\:tana + {tan}^{2}a = \dfrac{sina}{cosa} + \dfrac{cosa}{sina}$

$\rm :\longmapsto\:tana + {tan}^{2}a = tana + cota$

$\rm :\longmapsto\: {tan}^{2}a = cota$

$\rm :\longmapsto\: {tan}^{2}a = \dfrac{1}{tana}$

$\rm :\longmapsto\: {tan}^{3}a = 1$

$\rm\implies \:tana = 1$

$\rm\implies \:tana = tan\dfrac{\pi}{4}$

We know,

$\boxed{\tt{ tanx = tany \: \rm\implies \:\sf x = n\pi + y \: \forall \: n \in \: Z}}$

So, using this identity, we get

$\purple{\bf\implies \:a = n\pi + \dfrac{\pi}{4} \: \: \: \: \forall \: n \in \: Z}$

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MORE TO KNOW

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}$