Question

if cos4a=sin2a then the value of tan 4a is​

Answer 1

Step-by-step explanation:

tan

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Answer 2

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{cos4a \: = \: sin2a} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{tan \: 4a}  \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

$\tt \:  \longrightarrow \: cos \: 4a = sin \: 2a$

$\tt \:  \longrightarrow \: cos \: 4a \: = \: cos \: (90 \: - \: 2a)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \red { \bf \:  \{\because \: cos \: (90 \: - \: x) \: = \: sinx \}}$

$\tt\implies \:4a \: = \: 90 \: - \: 2a$

$\tt \:  \longrightarrow \: 4a \: + \: 2a \: = \: 90$

$\tt \:  \longrightarrow \: 6a \: = \: 90$

$\tt \:  \longrightarrow \: a \: = \: \dfrac{ \cancel{90} \: \: ^{15} }{ \cancel6}$

$\tt\implies \: \boxed{ \pink{\tt \:  a \: = \: 15}}$

$\begin{gathered}\bf\red{Now,}\end{gathered}$

$\tt \:  \longrightarrow \: \therefore \: tan \: 4a$

$\tt \:  \longrightarrow \: tan (\: 4 \times 15)$

$\tt \:  \longrightarrow \: tan \: 60$

$\tt \:  \longrightarrow \: \sqrt{3}$

$\tt\implies \: \boxed {\purple{\bf \:  tan \: 4a \: = \: \sqrt{3} }}$

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Additional Information:-

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

◇ Reciprocal Identities

The Reciprocal Identities are given as:

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

◇ Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

◇ Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

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