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Principal Solution of Trigonometric Equation

Prove that : ...

Question

Prove that :
${cos}^{2} θ (cos e c^{2} θ - {cot}^{2} θ) = {cos}^{2} θ .$

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Solution

L.H.S
$\begin{matrix} {cos}^{2} θ (c o {sec}^{2} θ - {cot}^{2} θ) = {cos}^{2} θ [c o {sec}^{2} θ - {cot}^{2} θ = 1] H e n c e, p r o v e d . \end{matrix}$

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\frac{{sin}^{2} θ}{{cos}^{2} θ} + \frac{{cos}^{2} θ}{{sin}^{2} θ} = {sec}^{2} θ - {cosec}^{2} θ - 2

s e c^{2} θ - c o t^{2} (90^{\circ} - θ) = c o s^{2} (90^{\circ} - θ) + c o s^{2} θ .

Q. Prove the following trigonometric identities.

(cosecθ + sinθ) (cosecθ − sinθ) = cot²θ + cos²θ

Q. Prove each of the following identities:

(i) \cot^{2} θ - \frac{1}{\sin^{2} θ} = - 1 (ii) \tan^{2} θ - \frac{1}{\cos^{2} θ} = - 1 (iii) \cos^{2} θ + \frac{1}{(1 + \cot^{2} θ)} = 1

\frac{{cos}^{2} θ}{{cot}^{2} θ} - {cos}^{2} θ = 3

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Principal Solution of Trigonometric Equation

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