Question

Hey Public,

If cosec theta = 41/40, find the value of cot theta and sec theta.

Let's Do copy paste.


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Answer 1

Answer:

hope this helps you dear

Answer 2

S O L U T I O N :

$\underline{\bf{Given\::}}$

cosec Ф = 41/40

$\underline{\bf{Explanation\::}}$

Firstly, attachment a figure of right angled triangle according to the given question.

As we know that,

$\boxed{\bf{cosec\:\theta = \frac{Hypotenuse}{Perpendicular} }}$

$\mapsto\tt{cosec\:\theta = \dfrac{41}{40} =\dfrac{BC}{AC} }$

A/q

$\underline{\mathcal{\red{USING\:\:BY\:\:PYTHAGORAS\:\:THEOREM\::}}}$

$\mapsto\sf{(Hypotenuse)^{2} = (Base)^{2} + (Perpendicular)^{2}}$

$\mapsto\sf{(BC)^{2} = (AB)^{2} + (AC)^{2}}$

$\mapsto\sf{(41)^{2} = (AB)^{2} + (40)^{2}}$

$\mapsto\sf{1681= (AB)^{2} + 1600}$

$\mapsto\sf{(AB)^{2} = 1681 - 1600}$

$\mapsto\sf{(AB)^{2} = 81}$

$\mapsto\sf{AB =\sqrt{81} }$

$\mapsto\sf{AB =9\:unit}$

Now,

$\boxed{\bf{cot\:\theta = \frac{Base}{Perpendicular} }}$

$\mapsto\tt{cot \:\theta = \dfrac{AB}{AC} }$

$\mapsto\tt{cot \:\theta = \dfrac{9}{40} }$

&

$\boxed{\bf{sec\:\theta = \frac{Hypotenuse}{Base} }}$

$\mapsto\tt{sec \:\theta = \dfrac{BC}{AB} }$

$\mapsto\tt{sec \:\theta = \dfrac{41}{9} }$

Thus,

The value of cot Ф & sec Ф will be 9/40 & 41/9 .