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Pythagoras Theorem

If θ=7/8, eva...

Question

If cot θ =, evaluate

(i) (ii) cot² θ

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Solution

Let us consider a right triangle ABC, right-angled at point B.

If BC is 7k, then AB will be 8k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

= (8k)² + (7k)²

= 64k² + 49k²

= 113k²

AC =

(i)

(ii) cot² θ = (cot θ)² = =

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Similar questions

{cot}^{2} θ = \frac{7}{8}

0 < θ < 90^{o}

, then the value of

\frac{(1 + sin θ) (1 - sin θ)}{(1 + cos θ) (1 - cos θ)}

If $cot θ = \frac{7}{8}$ , evaluate: $\frac{(1 + sin θ) (1 - sin θ)}{(1 + cos θ) (1 - cos θ)}$

Q. Prove each of the following identities:

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

Q. Prove the following trigonometric identities.

(i)

\sqrt{\frac{\sec θ - 1}{\sec θ + 1}} + \sqrt{\frac{\sec θ + 1}{\sec θ - 1}} = 2 cosec θ

\sqrt{\frac{1 + \sin θ}{1 - \sin θ}} + \sqrt{\frac{1 - \sin θ}{1 + \sin θ}} = 2 \sec θ

\sqrt{\frac{1 + \cos θ}{1 - \cos θ}} + \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} = 2 cosec θ

\frac{\sec θ - 1}{\sec θ + 1} = {(\frac{\sin θ}{1 + \cos θ})}^{2}

\frac{\sin θ + 1 - \cos θ}{\cos θ - 1 + \sin θ} = \frac{1 + \sin θ}{\cos θ}

sin θ (1 + sin θ) + cos θ (1 + cos θ) = x

sin θ (1 - sin θ) + cos θ (1 - cos θ) = y

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