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Prove that (2(pi + theta) (3pi)/(2) + theta)2((3pi)/(2) - theta) 2(pi - theta) (2pi - theta) =

Mathematics

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Here are the proofs for both trigonometric identities.

Part i· 5 steps

Prove that sin⁡2(π+θ)tan⁡(3π2+θ)cot⁡2(3π2−θ)cos⁡2(π−θ)csc⁡(2π−θ)=cos⁡θ\frac{\sin^2(\pi + \theta) \tan(\frac{3\pi}{2} + \theta)}{\cot^2(\frac{3\pi}{2} - \theta) \cos^2(\pi - \theta) \csc(2\pi - \theta)} = \cos\theta

  1. Simplify each trigonometric term using quadrant rules.

    • sin⁡(π+θ)=−sin⁡θ  ⟹  sin⁡2(π+θ)=(−sin⁡θ)2=sin⁡2θ\sin(\pi + \theta) = -\sin\theta \implies \sin^2(\pi + \theta) = (-\sin\theta)^2 = \sin^2\theta
    • tan⁡(3π2+θ)=−cot⁡θ\tan(\frac{3\pi}{2} + \theta) = -\cot\theta
    • cot⁡(3π2−θ)=tan⁡θ  ⟹  cot⁡2(3π2−θ)=tan⁡2θ\cot(\frac{3\pi}{2} - \theta) = \tan\theta \implies \cot^2(\frac{3\pi}{2} - \theta) = \tan^2\theta
    • cos⁡(π−θ)=−cos⁡θ  ⟹  cos⁡2(π−θ)=(−cos⁡θ)2=cos⁡2θ\cos(\pi - \theta) = -\cos\theta \implies \cos^2(\pi - \theta) = (-\cos\theta)^2 = \cos^2\theta
    • csc⁡(2π−θ)=−csc⁡θ\csc(2\pi - \theta) = -\csc\theta

  2. Substitute the simplified terms into the Left Hand Side (L.H.S.).

    L.H.S.=sin⁡2θ(−cot⁡θ)tan⁡2θ(cos⁡2θ)(−csc⁡θ)\text{L.H.S.} = \frac{\sin^2\theta (-\cot\theta)}{\tan^2\theta (\cos^2\theta) (-\csc\theta)}

  3. Simplify the expression.

    L.H.S.=−sin⁡2θcot⁡θ−tan⁡2θcos⁡2θcsc⁡θ\text{L.H.S.} = \frac{-\sin^2\theta \cot\theta}{-\tan^2\theta \cos^2\theta \csc\theta}

    The negative signs cancel out:

    L.H.S.=sin⁡2θcot⁡θtan⁡2θcos⁡2θcsc⁡θ\text{L.H.S.} = \frac{\sin^2\theta \cot\theta}{\tan^2\theta \cos^2\theta \csc\theta}

  4. Express cot⁡θ\cot\theta, tan⁡θ\tan\theta, and csc⁡θ\csc\theta in terms of sin⁡θ\sin\theta and cos⁡θ\cos\theta.

    Recall that cot⁡θ=cos⁡θsin⁡θ\cot\theta = \frac{\cos\theta}{\sin\theta}, tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}, and csc⁡θ=1sin⁡θ\csc\theta = \frac{1}{\sin\theta}.

    L.H.S.=sin⁡2θ(cos⁡θsin⁡θ)(sin⁡θcos⁡θ)2cos⁡2θ(1sin⁡θ)\text{L.H.S.} = \frac{\sin^2\theta \left(\frac{\cos\theta}{\sin\theta}\right)}{\left(\frac{\sin\theta}{\cos\theta}\right)^2 \cos^2\theta \left(\frac{1}{\sin\theta}\right)} L.H.S.=sin⁡2θcos⁡θsin⁡θsin⁡2θcos⁡2θcos⁡2θ1sin⁡θ\text{L.H.S.} = \frac{\sin^2\theta \frac{\cos\theta}{\sin\theta}}{\frac{\sin^2\theta}{\cos^2\theta} \cos^2\theta \frac{1}{\sin\theta}}

  5. Cancel common terms.

    In the numerator: sin⁡2θcos⁡θsin⁡θ=sin⁡θcos⁡θ\sin^2\theta \frac{\cos\theta}{\sin\theta} = \sin\theta \cos\theta.
    In the denominator: sin⁡2θcos⁡2θcos⁡2θ1sin⁡θ=sin⁡2θ1sin⁡θ=sin⁡θ\frac{\sin^2\theta}{\cos^2\theta} \cos^2\theta \frac{1}{\sin\theta} = \sin^2\theta \frac{1}{\sin\theta} = \sin\theta.

    L.H.S.=sin⁡θcos⁡θsin⁡θ\text{L.H.S.} = \frac{\sin\theta \cos\theta}{\sin\theta} L.H.S.=cos⁡θ\text{L.H.S.} = \cos\theta

    This is equal to the Right Hand Side (R.H.S.).
    Therefore, sin⁡2(π+θ)tan⁡(3π2+θ)cot⁡2(3π2−θ)cos⁡2(π−θ)csc⁡(2π−θ)=cos⁡θ\boxed{\frac{\sin^2(\pi + \theta) \tan(\frac{3\pi}{2} + \theta)}{\cot^2(\frac{3\pi}{2} - \theta) \cos^2(\pi - \theta) \csc(2\pi - \theta)} = \cos\theta}.

Part ii· 4 steps

Prove that cos⁡(90∘+θ)sec⁡(−θ)tan⁡(180∘−θ)sec⁡(360∘−θ)sin⁡(180∘+θ)cot⁡(90∘−θ)=−1\frac{\cos(90^\circ + \theta) \sec(-\theta) \tan(180^\circ - \theta)}{\sec(360^\circ - \theta) \sin(180^\circ + \theta) \cot(90^\circ - \theta)} = -1

  1. Simplify each trigonometric term using quadrant rules.

    • cos⁡(90∘+θ)=−sin⁡θ\cos(90^\circ + \theta) = -\sin\theta
    • sec⁡(−θ)=sec⁡θ\sec(-\theta) = \sec\theta (since sec⁡\sec is an even function)
    • tan⁡(180∘−θ)=−tan⁡θ\tan(180^\circ - \theta) = -\tan\theta
    • sec⁡(360∘−θ)=sec⁡θ\sec(360^\circ - \theta) = \sec\theta
    • sin⁡(180∘+θ)=−sin⁡θ\sin(180^\circ + \theta) = -\sin\theta
    • cot⁡(90∘−θ)=tan⁡θ\cot(90^\circ - \theta) = \tan\theta

  2. Substitute the simplified terms into the Left Hand Side (L.H.S.).

    L.H.S.=(−sin⁡θ)(sec⁡θ)(−tan⁡θ)(sec⁡θ)(−sin⁡θ)(tan⁡θ)\text{L.H.S.} = \frac{(-\sin\theta)(\sec\theta)(-\tan\theta)}{(\sec\theta)(-\sin\theta)(\tan\theta)}

  3. Simplify the expression.

    Multiply the terms in the numerator: (−sin⁡θ)(sec⁡θ)(−tan⁡θ)=sin⁡θsec⁡θtan⁡θ(-\sin\theta)(\sec\theta)(-\tan\theta) = \sin\theta \sec\theta \tan\theta.
    Multiply the terms in the denominator: (sec⁡θ)(−sin⁡θ)(tan⁡θ)=−sin⁡θsec⁡θtan⁡θ(\sec\theta)(-\sin\theta)(\tan\theta) = -\sin\theta \sec\theta \tan\theta.

    L.H.S.=sin⁡θsec⁡θtan⁡θ−sin⁡θsec⁡θtan⁡θ\text{L.H.S.} = \frac{\sin\theta \sec\theta \tan\theta}{-\sin\theta \sec\theta \tan\theta}

  4. Cancel common terms.

    Assuming sin⁡θ≠0\sin\theta \neq 0, sec⁡θ≠0\sec\theta \neq 0, and tan⁡θ≠0\tan\theta \neq 0:

    L.H.S.=−1\text{L.H.S.} = -1

    This is equal to the Right Hand Side (R.H.S.).
    Therefore, cos⁡(90∘+θ)sec⁡(−θ)tan⁡(180∘−θ)sec⁡(360∘−θ)sin⁡(180∘+θ)cot⁡(90∘−θ)=−1\boxed{\frac{\cos(90^\circ + \theta) \sec(-\theta) \tan(180^\circ - \theta)}{\sec(360^\circ - \theta) \sin(180^\circ + \theta) \cot(90^\circ - \theta)} = -1}.

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