Question

plz solve this problem.. ​

Answer 1

To prove -

( cos theta - 2 cos ³ theta )/(2 sin³ theta - sin theta ) = cot theta

Proof -

( cos theta - 2 cos ³ theta )/(2 sin³ theta - sin theta )

> cos theta [ 1 - 2 cos² theta ]/ sin theta [ 2 sin² theta - 1 ]

> tan theta × [ 1 - 2 cos² theta ]/[ 2 sin² theta - 1]

> tan theta × [ sin² theta + cos² theta - 2 cos² theta ]/[ 2 sin² theta - sin² theta - cos²theta ]

> tan theta × [ sin ² theta - cos ²theta ]/[ sin² theta - cos² theta ]

> tan theta

Hence Disproved !

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Additional Information -

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

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Answer 2

Step-by-step explanation:

cos⊙-2cos^3⊙/

2sin^3⊙-sin⊙ =cot⊙

consider L.H.S

cos⊙(1-2cos^2⊙)/

sin⊙(2sin^2⊙-1)

=cos⊙.cos2⊙/

sin⊙.cos2⊙ cos2⊙ get cancelled

=cos⊙/sin⊙

=cot⊙

HENCE PROOVED.