Question

if sin theth a= x and sec thetha = y then find the value of cot thetha

Answer 1

Answer:

It is stated that sin θ A is x whereas the sec θ A is y. So, from this we have to find out the value of cot θ A.

So, For that we are going to use some trigonometry function.

We know that,

${\bullet \: \:{\boxed {\bf {Tan \: \theta = \dfrac{Sin \: \theta}{Cos \: \theta}}}}}$

Whereas, cos is it's reciprocal i.e,

${\bullet \: \: {\boxed{\bf {Cot \: \theta = \dfrac{Cos \: \theta}{Sin \: \theta}}}}}$

Also,

${\bullet \: \: {\boxed{\bf {Cos \: \theta = \dfrac{1}{Sec \: \theta}}}}}$

Let's solve it :

From Given :-

• Sin θ A = x

• Sec θ A = y

Then ,

${\bullet \: \: {\bf {Cos \: \theta A = \dfrac{1}{y}}}}$

Plugging the respective values in it, we get :

$\dashrightarrow\:\:\sf Cot \: \theta = \dfrac{Cos \: \theta}{Sin \: \theta}$

$\dashrightarrow\:\:\sf Cot \: \theta = \dfrac{ \dfrac{1}{y}}{x}$

$\dashrightarrow\:\:\sf Cot \: \theta = \dfrac{1}{y} \times x$

$\dashrightarrow\:\:\sf Cot \: \theta = \dfrac{1}{yx}$

Hence,

$\dashrightarrow\:\: \underline{ \boxed{\sf Required \: answer : {\red{\dfrac{1}{xy}}}}}$

_____________________

Related Data :

• Sin A = 1/Cosec A

• Cos A = 1/Sec A

• Tan = 1/ Cot A

• Cot A = 1/Tan A

• Cosec A = 1/Sin A

• Tan A = Sin A/Cos A

• Cot A = Cos A/Sin A

Trigonometry Identities :

• sin² A + cos² A = 1

• sec² A = 1 + tan² A

• cosec² A = 1 + cot² A

Answer 2

Given : The $\sin\theta$ is x & $\sec\theta$ is y .

Exigency to find : The Value of $\cot\theta$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\dag\:\:\it{Given\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{ \sin \theta = x \:\&\:\sec\theta =y }\bigg\rgroup$

⠀⠀⠀⠀⠀Finding Value of $\cot\theta$ :

$\dag\:\:\it{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{ \cos \theta = \dfrac{1}{\sec\theta } }\bigg\rgroup$

⠀⠀⠀⠀⠀Here $\sec\theta$ = y .

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \longmapsto \sf \cos \theta = \dfrac{1}{y}$

⠀⠀⠀⠀⠀&

$\dag\:\:\it{\:We\:know\:that\:to\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{ \cot \theta = \dfrac{\cos\theta}{\sin\theta } }\bigg\rgroup$

⠀⠀⠀⠀⠀Here $\sin\theta$ is x & $\cos\theta\:\:is\:\:\dfrac{1}{y}$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \longmapsto \sf \cot \theta = \dfrac{\dfrac{1}{y}}{x}$

$\qquad \longmapsto \sf \cot \theta = \dfrac{1}{x\times y }$

$\qquad \longmapsto \bf\bigg( \cot \theta = \dfrac{1}{xy } \bigg) \qquad \longrightarrow \:\:Required \:AnswEr \:$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:The\:Value \:of\:\cot\theta \:is\:\bf{\dfrac{1}{xy}}}}}$

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❍ Some Basic Formulas of Trigonometry is given by :

$\boxed { \begin{array}{c c} \\ \dag \qquad \large {\underline {\bf{ Some \:Basic\:Formulas \:For\:Trigonometry \::}}}\\\\ \sf{ In \:a \:Right \:Angled \: Triangle-:} \\\\ \sf {\star Sin \theta = \dfrac{1}{Cosec \theta }} \\\\ \sf{ \star \cos \theta = \dfrac{ 1 }{Sec\theta}}\\\\ \sf{\star \tan \theta = \dfrac{1}{Cot \theta }}\\\\ \sf{\star \cot \theta = \dfrac{1}{Tan\theta}} \\\\ \sf {\star Sec \theta = \dfrac{1}{Cos \theta }} \\\\ \sf{\star \cosec \theta = \dfrac{1}{Sin \theta }}\\\\ \sf{\star \cot \theta = \dfrac{Cos\theta}{Sin\theta}} \end{array}}$

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