Question

prove the identity of the given sum​

Answer 1

Answer:

answer for the given problem is given

Answer 2

To Prove:-

$\tt \to \:\dfrac{sin \: \theta - 2 {sin}^{3}\: \theta }{2 {cos}^{3} \: \theta - cos\: \theta} = tan\: \theta$

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$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$\large \boxed{ \tt \:  \red{ ⟼ (1) \: {sin}^{2} \: \theta + {cos}^{2} \: \theta = 1}}$

$\large \boxed{ \tt \:  \red{ ⟼ (2) \: \:\dfrac{sin\: \theta}{cos\: \theta} = tan\: \theta}}$

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

$\tt \: \to \:Consider \: LHS \: \:\dfrac{sin \: \theta - 2 {sin}^{3}\: \theta }{2 {cos}^{3} \: \theta - cos\: \theta}$

$\tt \:  ⟼ \: = \: \dfrac{sin\: \theta(1 - 2 {sin}^{2}\: \theta )}{cos\: \theta(2 {cos}^{2}\: \theta - 1) }$

$\tt \:  ⟼ \: = \: tan\: \theta \: \times \dfrac{1 - {2(1 - cos}^{2} \: \theta)}{ {2cos}^{2}\: \theta - 1 }$

$\tt \:  ⟼ \: = \: tan\: \theta \times \dfrac{1 - 2 + 2 {cos}^{2} \: \theta}{ {2cos}^{2} \: \theta - 1}$

$\tt \:  ⟼ \: = \: tan \: \theta \times \dfrac{ \cancel{ {2cos}^{2} \: \theta - 1}}{ \cancel{{2cos}^{2} \: \theta - 1}}$

$\tt \:  ⟼ \: = \: tan\: \theta$

$\tt \:  ➦ LHS = RHS$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

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$\large \red{\tt \: ⟼ Explore \: \: more } ✍$

Additional Information:-

Additional Information:- Relationship between sides and T ratios

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

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

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