Question

If 1/sin +1/tan =9,then find the value of cot -cosec? with explanation ​

Answer 1

Step-by-step explanation:

HERE IS YOUR ANSWER

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take LCM

WE GET

$here \: it \: is \: given \: that \: \\ \cot( \alpha ) + \csc( \alpha ) = 9 \\ \\ and \: we \: know \: that \: \\ \\ { \csc( \alpha ) }^{2} - { \cot( \alpha ) }^{2} = 1 \\ \\ \\ \\ \\ so \: required \: answer \: = \frac{1}{9}$

Answer 2

$- \dfrac{1}{9}$

Step-by-step explanation:

Given:

$\dfrac{1}{ \sin } + \dfrac{1}{ \tan } = 9$

To find:

$\cot - \cosec$

Solution:

$\dfrac{1}{ \sin } + \dfrac{1}{ \tan } = 9 \\ \\ ⟹ \cosec + \cot \: = 9 \: \: \: .......(i)$

As We know,

${ \cosec}^{2} - \cot {}^{2} = 1 \\ \\ ⟹ ({ \cosec}^{} + \cot {}^{})({ \cosec}^{} - \cot {}^{}) = 1 \\ \\ ⟹9({ \cosec}^{} - \cot {}^{}) = 1 \: \: \: \: \: [ \: From \: \: i ] \\ \\⟹ ({ \cosec}^{} - \cot {}^{}) = \frac{1}{9} \\ \\ ⟹- ({ \cosec}^{} - \cot {}^{}) = - \frac{1}{9} \\ \\⟹ \cot {}^{}{ - \cosec}^{} = - \frac{1}{9}$

Hence, the required answer is - 1/9 .

Some important trigonometry Rules:

sinø . cosecø = 1

cosø . secø = 1

tanø . cotø = 1

sin²ø + cos²ø = 1

cosec²ø - cot² = 1

sec²ø - tan²ø = 1