Question

prove that cos square theta (1 + tan square theta) + sin square theta (1 + cot square theta) =2

Answer 1

Answer:

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Answer 2

Question:

Prove that:

cos²θ (1 + tan²θ) + sin²θ (1 + cot²θ) =2

Answer:

RHS:

$= > { \cos }^{2} \theta(1 + { \tan}^{2} \theta) + { \sin }^{2} \theta(1 + { \cot }^{2} \theta) \\ \\ = > { \cos }^{2} \theta(1 + \frac{{ \sin}^{2} \theta}{{ \cos }^{2} \theta} ) + { \sin }^{2} \theta(1 + \frac{{ \cos }^{2} \theta}{{ \sin }^{2} \theta} ) \\ \\ = >({ \cos }^{2} \theta + { \cos }^{2} \theta \times \frac{{ \sin}^{2} \theta}{{ \cos }^{2} \theta} ) + ({ \sin }^{2} \theta + { \sin}^{2} \theta \times \frac{{ \cos}^{2} \theta}{{ \sin }^{2} \theta} ) \\ \\ = > ( { \cos}^{2} \theta + { \sin}^{2} \theta) + ({ \sin}^{2} \theta + { \cos}^{2} \theta) \\ \\ = > 1 + 1 \\ \\ = > 1$

LHS:

=> 2

LHS = RHS

Hence proved