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Complementary Trigonometric Ratios

If θ=78, eval...

Question

If $\cot θ = \frac{7}{8}$ , evaluate :

(i) $\frac{(1 + \sin θ) (1 - \sin θ)}{(1 + \cos θ) (1 - \cos θ)}$
(ii) cot² θ

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Solution

(i) Given:

To evaluate:

…… (1)

We know the following formula

By applying the above formula in the numerator of equation (1) ,

We get,

… (Where a = 1 and b = )

…… (2)

Similarly,

By applying formula in the denominator of equation (1) ,

We get,

… (Where a = 1 and b = )

…… (3)

Substituting the value of numerator and denominator of equation (1) ,from equation (2) and(3)

Therefore,

…… (4)

Since,

Therefore,

Putting the value of and in Equation (4)

We get,

We know that,

Since,

Therefore,

Answer:

(ii) Given:

To evaluate:

Squaring on both sides,

We get,

Answer:

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Similar questions

cot θ = \frac{7}{8},

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

{cot}^{2} θ

\frac{1 - sinθ}{1 + sinθ} = (secθ - tanθ)^{2}

\frac{1 + cosθ}{1 - cosθ} = (cosecθ + cotθ)^{2}

{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)}

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 + tan θ) + c o s θ (1 + cot θ) = csc θ + sec θ

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