Question

If sec theta + tan theta = p
then find the value of cosec theta​

enjoy​

Answer 1

Answer:

Given :

sec Ф + tan Ф = р

Solution :

$sec \: \alpha + tan \: \alpha = p \\ \\ {sec}^{2} \alpha - {tan}^{2} \alpha = 1 \\ = > (sec \: \alpha - tan \: \alpha ) \times p = 1 \\ = > sec \: \alpha - tan \: \alpha = \frac{1}{p}$

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Adding the above 2 equations,

$2sec \: \alpha = p + \frac{1}{p} \\ = > sec \: \alpha = \frac{ {p}^{2} + 1 }{2p}$

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$cos \: \alpha = \frac{1}{sec \: \alpha } \\ = > cos \: \alpha = \frac{2p}{ {p}^{2} + 1 }$

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${sin}^{2} \: \alpha = 1 - {cos}^{2} \alpha \\ = > {sin}^{2} \: \alpha = 1 - \frac{2p}{ {p}^{2} + 1 } \\ = > sin \: \alpha = \sqrt{ \frac{ {(p - 1)}^{2} }{ {p}^{2} + 1} } \\ = > sin \: \alpha = \frac{p - 1}{ \sqrt{ {p}^{2} + 1 } }$

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$cosec \: \alpha = \frac{1}{sin \: \alpha } \\ \\ = > cosec \: \alpha = \frac{ \sqrt{ {p}^{2} + 1 } }{p - 1}$

Answer 2

Given :

$\rightarrow \sf {sec ~Ф + tan~ Ф = р}$

Solution :

$sec \: \alpha + tan \: \alpha = p \\ \\ {sec}^{2} \alpha - {tan}^{2} \alpha = 1 \\ = > (sec \: \alpha - tan \: \alpha ) \times p = 1 \\ = > sec \: \alpha - tan \: \alpha = \frac{1}{p}$

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Adding the above 2 equations,

$2sec \: \alpha = p + \frac{1}{p} \\ = > sec \: \alpha = \frac{ {p}^{2} + 1 }{2p}$

__________

$cos \: \alpha = \frac{1}{sec \: \alpha } \\ = > cos \: \alpha = \frac{2p}{ {p}^{2} + 1 }$

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${sin}^{2} \: \alpha = 1 - {cos}^{2} \alpha \\ = > {sin}^{2} \: \alpha = 1 - \frac{2p}{ {p}^{2} + 1 } \\ = > sin \: \alpha = \sqrt{ \frac{ {(p - 1)}^{2} }{ {p}^{2} + 1} } \\ = > sin \: \alpha = \frac{p - 1}{ \sqrt{ {p}^{2} + 1 } }$

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$cosec \: \alpha = \frac{1}{sin \: \alpha } \\ \\ = > cosec \: \alpha = \frac{ \sqrt{ {p}^{2} + 1 } }{p - 1}$