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Prove the trigonometric identities: 1) tan θ sin θ + cos θ = sec θ 2) 1/tan θ + tan θ = 1/(sin θ cos θ)

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: Prove tan⁡θsin⁡θ+cos⁡θ=sec⁡θ\tan \theta \sin \theta + \cos \theta = \sec \theta

  1. Start with the Left Hand Side (LHS).

    LHS=tan⁡θsin⁡θ+cos⁡θ\text{LHS} = \tan \theta \sin \theta + \cos \theta

  2. Express tan⁡θ\tan \theta as sin⁡θcos⁡θ\frac{\sin \theta}{\cos \theta}.

    LHS=sin⁡θcos⁡θ⋅sin⁡θ+cos⁡θ\text{LHS} = \frac{\sin \theta}{\cos \theta} \cdot \sin \theta + \cos \theta LHS=sin⁡2θcos⁡θ+cos⁡θ\text{LHS} = \frac{\sin^2 \theta}{\cos \theta} + \cos \theta

  3. Find a common denominator and combine the terms.

    LHS=sin⁡2θ+cos⁡2θcos⁡θ\text{LHS} = \frac{\sin^2 \theta + \cos^2 \theta}{\cos \theta}

  4. Use the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1.

    LHS=1cos⁡θ\text{LHS} = \frac{1}{\cos \theta}

  5. Express in terms of sec⁡θ\sec \theta.

    LHS=sec⁡θ\text{LHS} = \sec \theta

    Since LHS = RHS, the identity is proven.
    Proven\boxed{\text{Proven}}

    : Prove 1tan⁡θ+tan⁡θ=1sin⁡θcos⁡θ\frac{1}{\tan \theta} + \tan \theta = \frac{1}{\sin \theta \cos \theta}

  6. Start with the Left Hand Side (LHS).

    LHS=1tan⁡θ+tan⁡θ\text{LHS} = \frac{1}{\tan \theta} + \tan \theta

  7. Express 1tan⁡θ\frac{1}{\tan \theta} as cos⁡θsin⁡θ\frac{\cos \theta}{\sin \theta} and tan⁡θ\tan \theta as sin⁡θcos⁡θ\frac{\sin \theta}{\cos \theta}.

    LHS=cos⁡θsin⁡θ+sin⁡θcos⁡θ\text{LHS} = \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta}

  8. Find a common denominator, which is sin⁡θcos⁡θ\sin \theta \cos \theta.

    LHS=cos⁡2θsin⁡θcos⁡θ+sin⁡2θsin⁡θcos⁡θ\text{LHS} = \frac{\cos^2 \theta}{\sin \theta \cos \theta} + \frac{\sin^2 \theta}{\sin \theta \cos \theta}

  9. Combine the terms.

    LHS=cos⁡2θ+sin⁡2θsin⁡θcos⁡θ\text{LHS} = \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta}

  10. Use the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1.

    LHS=1sin⁡θcos⁡θ\text{LHS} = \frac{1}{\sin \theta \cos \theta}

    Since LHS = RHS, the identity is proven.
    Proven\boxed{\text{Proven}}

    : If sin⁡A=35\sin A = \frac{3}{5}, prove that tan⁡A+1cos⁡A=2\tan A + \frac{1}{\cos A} = 2 or −2-2.

  11. Given sin⁡A=35\sin A = \frac{3}{5}.

    Use the identity sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1 to find cos⁡A\cos A.

    (35)2+cos⁡2A=1\left(\frac{3}{5}\right)^2 + \cos^2 A = 1 925+cos⁡2A=1\frac{9}{25} + \cos^2 A = 1 cos⁡2A=1−925=1625\cos^2 A = 1 - \frac{9}{25} = \frac{16}{25} cos⁡A=±1625=±45\cos A = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5}

  12. Consider two cases for cos⁡A\cos A.

    • Case 1: cos⁡A=45\cos A = \frac{4}{5} (A is in Quadrant I)

      • Calculate tan⁡A\tan A: tan⁡A=sin⁡Acos⁡A=3/54/5=34\tan A = \frac{\sin A}{\cos A} = \frac{3/5}{4/5} = \frac{3}{4}
      • Substitute values into the expression tan⁡A+1cos⁡A\tan A + \frac{1}{\cos A}: tan⁡A+1cos⁡A=34+14/5=34+54=84=2\tan A + \frac{1}{\cos A} = \frac{3}{4} + \frac{1}{4/5} = \frac{3}{4} + \frac{5}{4} = \frac{8}{4} = 2

    • Case 2: cos⁡A=−45\cos A = -\frac{4}{5} (A is in Quadrant II)

      • Calculate tan⁡A\tan A: tan⁡A=sin⁡Acos⁡A=3/5−4/5=−34\tan A = \frac{\sin A}{\cos A} = \frac{3/5}{-4/5} = -\frac{3}{4}
      • Substitute values into the expression tan⁡A+1cos⁡A\tan A + \frac{1}{\cos A}: tan⁡A+1cos⁡A=−34+1−4/5=−34−54=−84=−2\tan A + \frac{1}{\cos A} = -\frac{3}{4} + \frac{1}{-4/5} = -\frac{3}{4} - \frac{5}{4} = \frac{-8}{4} = -2

    Thus, tan⁡A+1cos⁡A\tan A + \frac{1}{\cos A} can be 2 or −2-2. Proven\boxed{\text{Proven}}

    **: If θ\theta is in the fourth quadrant and cos⁡θ=513\cos \theta = \frac{5}{13}, find the value of $\frac{13 \sin \theta + 5 \sec \theta}{5 \tan \theta

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