Question

If cosec theta +cot theta=k then write all trigonometric ratios at theta in terms of k.

Answer 1

Hope it helps.
Thankyou

Answer 2

The value of other trigonometric ratios in terms of k are

$\bf \red{sin \: \theta = \frac{2k}{1 + {k}^{2} }}$

$\bf \green{{cosec} \: \theta = \frac{1 + {k}^{2} }{2 k} }$

$\bf \red{cos\: \theta = \frac{1 - {k}^{2} }{1 + {k}^{2} } }$

$\bf \green{sec\: \theta = \frac{1 + {k}^{2} }{1 - {k}^{2} } }$

$\bf \red{tan \: \theta = \frac{2k}{1 - {k}^{2} } }$

$\bf \green{cot \: \theta = \frac{1 - {k}^{2} }{2k} }$

Given:

• $cosec( \theta) + \cot(\theta) = k ...eq1$

To find:

• Write all trigonometric ratios at theta in terms of k.

Solution:

Trigonometric / algebraic identity to be used:

${cosec}^{2} \theta - {cot}^{2} \theta = 1$

${a}^{2} - {b}^{2} = (a + b)(a - b)$

$(a+b)^2={a}^{2} +2ab+{b}^{2}$

$(a-b)^2={a}^{2} -2ab+{b}^{2}$

Step 1:

Apply algebraic identity on trigonometric identity.

${cosec}^{2} \theta - {cot}^{2} \theta = 1$

or

$({cosec} \: \theta - {cot} \: \theta)({cosec} \: \theta + {cot }\: \theta) = 1$

put the value from given terms.

$({cosec} \:\theta - {cot}\: \theta)k = 1$

or

$({cosec} \:\theta - {cot}\:\theta)= \frac{1}{k} ...eq2$

Step 2:

Add both equations.

${cosec}\:\theta - {cot}\:\theta= \frac{1}{k} \\ {cosec}\:\theta + {cot}\:\theta = k \\ - - - - - - - - - \\ 2 \: {cosec}\:\theta = \frac{1}{k} + k \\ - - - - - - - - -$

or

$\bf \green{{cosec} \: \theta = \frac{1 + {k}^{2} }{2 k} }$

We know that

$\bf sin \: \theta = \frac{1}{cosec \: \theta}$

so,

$\bf \red{sin \: \theta = \frac{2k}{1 + {k}^{2} }}$

Step 3:

As,

$\bf cos\: \theta = \sqrt{1 - {sin}^{2} \theta}$

so,

$cos\: \theta = \sqrt{1 - { \left( \frac{2k}{1 + {k}^{2} } \right)}^{2} }$

or

$cos\: \theta = \sqrt{1 - { \left( \frac{4 {k}^{2} }{1 +2 {k}^{2} + {k}^{4} } \right)} }$

or

$cos\: \theta = \sqrt{ { \frac{1 +2 {k}^{2} + {k}^{4} - 4 {k}^{2} }{1 +2 {k}^{2} + {k}^{4} } } }$

or

$cos\: \theta = \sqrt{ { \frac{1 - 2 {k}^{2} + {k}^{4} }{1 +2 {k}^{2} + {k}^{4} } } }$

or

$cos\: \theta = \sqrt{ { \left( \frac{1 - {k}^{2} }{1 + {k}^{2} } \right)}^{2} }$

or

$\bf \red{cos\: \theta = \frac{1 - {k}^{2} }{1 + {k}^{2} } }$

As,

$\bf cos\: \theta = \frac{1}{sec\: \theta}$

Thus,

$\bf \green{sec\: \theta = \frac{1 + {k}^{2} }{1 - {k}^{2} } }$

Step 4:

As,

$\bf tan \: \theta = \frac{sin\: \theta}{cos\: \theta}$

so,

$tan \: \theta = \frac{2k}{1 + {k}^{2} } \times \frac{1 + {k}^{2} }{1 - {k}^{2} }$

or

$\bf \pink{tan \: \theta = \frac{2k}{1 - {k}^{2} } }$

As,

$\bf cot \: \theta = \frac{1}{tan \: \theta}$

so,

$\bf \pink{cot \: \theta = \frac{1 - {k}^{2} }{2k} }$

Thus,

All the trigonometric ratios have calculated in terms of k.

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