Question

if cosecA + cotA = 3. then find the value of cosA.pls solve on your copy and I will mark brainliest

Answer 1

Hey

$\frac{1 + \cos( \alpha ) }{ \sin( \alpha ) } = 3$

-> We'd want to get rid of either of the two to get a pretty, solvable Quadratic Equation

So, squaring both sides :

$1 + 2 \cos( \alpha ) + \cos {}^{2} ( \alpha ) = 9 \sin {}^{2} ( \alpha )$

Converting the sine to cosine ->

$5 \cos {}^{2} ( \alpha ) + \cos( \alpha ) - 4 = 0$

$(5 \cos( \alpha ) -4)( \cos( \alpha ) + 1) = 0$

$= > \cos( \alpha ) = \frac{4}{5} \: or \: ( - 1)$

Now, it's perhaps cool that

-> cos A = -1 ( which makes sine A = 0 and cosec A = ... well you know what )

So, it's perhaps ^^" Better we avoid things not in our hands !!

Only possible Solution is :

$\cos( \alpha ) = \frac{4}{5}$
____________________________________

Or

$\csc {}^{2} ( \alpha ) - \cot {}^{2} ( \alpha ) = 1$

And since, don't know when ^^" we have a beautiful Factorisation :->

$( \csc( \alpha ) + \cot( \alpha ) ) ( \csc( \alpha ) - \cot( \alpha ) ) = 1$

∆ Fill in the known value to get :->

$\csc( \alpha ) - \cot( \alpha ) = \frac{1}{3}$

Add something ->

$2 \csc( \alpha ) = \frac{10}{3}$

$\csc( \alpha ) = \frac{5}{3}$

$= > \cos( \alpha ) = \frac{4}{5}$

Choose your easy way :smile:

Answer 2

Answer:

Hey

\frac{1 + \cos( \alpha ) }{ \sin( \alpha ) } = 3

sin(α)

1+cos(α)

=3

-> We'd want to get rid of either of the two to get a pretty, solvable Quadratic Equation

So, squaring both sides :

1 + 2 \cos( \alpha ) + \cos {}^{2} ( \alpha ) = 9 \sin {}^{2} ( \alpha )1+2cos(α)+cos

2

(α)=9sin

2

(α)

Converting the sine to cosine ->

5 \cos {}^{2} ( \alpha ) + \cos( \alpha ) - 4 = 05cos

2

(α)+cos(α)−4=0

(5 \cos( \alpha ) -4)( \cos( \alpha ) + 1) = 0(5cos(α)−4)(cos(α)+1)=0

= > \cos( \alpha ) = \frac{4}{5} \: or \: ( - 1)=>cos(α)=

5

4

or(−1)

Now, it's perhaps cool that

-> cos A = -1 ( which makes sine A = 0 and cosec A = ... well you know what )

So, it's perhaps ^^" Better we avoid things not in our hands !!

Only possible Solution is :

\cos( \alpha ) = \frac{4}{5}cos(α)=

5

4

____________________________________

Or

\csc {}^{2} ( \alpha ) - \cot {}^{2} ( \alpha ) = 1csc

2

(α)−cot

2

(α)=1

And since, don't know when ^^" we have a beautiful Factorisation :->

( \csc( \alpha ) + \cot( \alpha ) ) ( \csc( \alpha ) - \cot( \alpha ) ) = 1(csc(α)+cot(α))(csc(α)−cot(α))=1

∆ Fill in the known value to get :->

\csc( \alpha ) - \cot( \alpha ) = \frac{1}{3}csc(α)−cot(α)=

3

1

Add something ->

2 \csc( \alpha ) = \frac{10}{3}2csc(α)=

3

10

\csc( \alpha ) = \frac{5}{3}csc(α)=

3

5

= > \cos( \alpha ) = \frac{4}{5}=>cos(α)=

5

4

Choose your easy way :smile: