Question

k²-1
k²+1
If cosec 0 + cot 0=k, then prove that cos e​

Answer 1

Question : -

If cosec x + cot x = k the prove that cos x = (k²-1)/(k²+1)

ANSWER

Given : -

cosec x + cot x = k

Required to prove : -

• cos x = (k²-1)/(k²+1)

Proof : -

cosec x + cot x = k

Consider the proof !

cos x = (k²-1)/(k²+1)

Consider the RHS Part

(k²-1)/(k²+1)

since,

• cosec x + cot x = k

([cosec x + cot x]² - 1)/([cosec x + cot x]² + 1)

Using the algebraic identity;

• (a+b)² = a²+b²+2ab

(cosec² x + cot² x + 2*cosec x*cot x - 1)/(cosec² x + cot² x + 2*cosec x*cot x + 1)

Here, in the numerator we are having "-1" let's perform some steps to remove it .

We know that;

• cosec² x - cot² x = 1
• cosec² x = 1 + cot² x ....(1)

This implies;

([1+cot² x]+ cot² x + 2*cosec x*cot x - 1)/(cosec² x + cot² x + 2*cosec x*cot x + 1)

(1+cot² x + cot² x + 2*cosec x*cot x - 1)/(cosec² x + cot² x + 2*cosec x*cot x + 1)

+1,-1 get's cancelled in numerator

(2cot² x + 2*cosec x*cot x)/(cosec² x + cot² x + 2*cosec x*cot x + 1)

Now,

In the denominator we need to remove '1'

So,

We know that;

• cosec² x - cot² x = 1
• cosec² x - 1 = cot² x ....(2)

(2cot² x + 2*cosec x*cot x)/(cosec² x + [cosec² x - 1] + 2*cosec x*cot x + 1)

(2cot² x + 2*cosec x*cot x)/(cosec² x + cosec² x - 1 + 2*cosec x*cot x + 1)

-1,+1 get's cancelled on in the denominator

(2cot² x + 2*cosec x*cot x)/(2cosec² x + 2*cosec x*cot x)

Since,

• cot x = (cos x)/(sin x)
• cosec x = (1)/(sin x)

(2[cos² x]/[sin² x] + 2*[1]/[sin x]*[cos x]/[sin x])/(2[1]/[sin² x] + 2*[1]/[sin x]*[cos x]/[sin x])

([2cos² x]/[sin² x] + [2cos x]/[sin² x])/([2]/[sin² x] + [2cos x]/[sin² x])

([2cos² x + 2 cos x]/[sin² x])/([2 + 2 cos x]/[sin² x])

([2cos² x + 2 cos x]/[sin² x]) ÷ ([2 + 2 cos x]/[sin² x])

([2cos² x + 2 cos x]/[sin² x]) x ([sin² x]/[2 + 2cos x])

sin² x get's cancelled on both numerator and denominator

(2cos² x + 2 cos x)/(2 + 2cos x)

Taking 2 cos x common in numerator & 2 common in denominator

(2cos x[cos x + 1])/(2[1 + cos x])

(2cos x[1 + cos x])/(2[1+cos x])

1+cos x get's cancelled on both numerator and denominator

(2cos x)/(2)

Cancelling 2 in both numerator and denominator

cos x

= RHS

LHS = RHS

Hence Proved !

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

• If cosec x + cot x = k the prove that cos x = (k²-1)/(k²+1)

★═════════════════★

♣ ɢɪᴠᴇɴ :

• cosec x + cot x = k

★═════════════════★

♣ ᴛᴏ ᴘʀᴏᴠᴇ :

• cos x = (k²-1)/(k²+1)

★═════════════════★

♣ ᴀɴꜱᴡᴇʀ :

cosec x + cot x = k

cos x =

cos x =

Now To Prove :

cos x =

Steps :

$\mathrm{Manipulating\:right\:side}:$

$\sf{\dfrac{\left(\csc \left(x\right)+\cot \left(x\right)\right)^2-1}{\left(\csc \left(x\right)+\cot \left(x\right)\right)^2+1}}$

$\mathrm{Use\:the\:following\:identity:}\:1=-\cot ^2\left(x\right)+\csc ^2\left(x\right)$

$\sf{=\dfrac{-\left(-\cot ^2\left(x\right)+\csc ^2\left(x\right)\right)+\left(\cot \left(x\right)+\csc \left(x\right)\right)^2}{\left(\cot \left(x\right)+\csc \left(x\right)\right)^2-\cot ^2\left(x\right)+\csc ^2\left(x\right)}}$

$\sf{Simplify\:\:\dfrac{-\left(-\cot ^2\left(x\right)+\csc ^2\left(x\right)\right)+\left(\cot \left(x\right)+\csc \left(x\right)\right)^2}{\left(\cot \left(x\right)+\csc \left(x\right)\right)^2-\cot ^2\left(x\right)+\csc ^2\left(x\right)}}:$

______________________

$\sf{\dfrac{-\left(-\cot ^2\left(x\right)+\csc ^2\left(x\right)\right)+\left(\cot \left(x\right)+\csc \left(x\right)\right)^2}{\left(\cot \left(x\right)+\csc \left(x\right)\right)^2-\cot ^2\left(x\right)+\csc ^2\left(x\right)}}$

$=\dfrac{-\left(\csc ^2\left(x\right)-\cot ^2\left(x\right)\right)+\left(\cot \left(x\right)+\csc \left(x\right)\right)^2}{2\cot \left(x\right)\csc \left(x\right)+2\csc ^2\left(x\right)}$

$=\dfrac{2\cot ^2\left(x\right)+2\cot \left(x\right)\csc \left(x\right)}{2\cot \left(x\right)\csc \left(x\right)+2\csc ^2\left(x\right)}$

$=\dfrac{2\cot \left(x\right)\left(\cot \left(x\right)+\csc \left(x\right)\right)}{2\cot \left(x\right)\csc \left(x\right)+2\csc ^2\left(x\right)}$

$=\dfrac{2\cot \left(x\right)\left(\cot \left(x\right)+\csc \left(x\right)\right)}{2\csc \left(x\right)\left(\cot \left(x\right)+\csc \left(x\right)\right)}$

$\sf{=\dfrac{\cot \left(x\right)}{\csc \left(x\right)}}$

______________________

$\bigstar\:\:\sf{\dfrac{-\left(-\cot ^{2}(x)+\csc ^{2}(x)\right)+(\cot (x)+\csc (x))^{2}}{(\cot (x)+\csc (x))^{2}-\cot ^{2}(x)+\csc ^{2}(x)}: \dfrac{\cot (x)}{\csc (x)}}\:\:\bigstar$

______________________

$\text { Express with sin, cos }$

$\sf{\dfrac{\cot \left(x\right)}{\csc \left(x\right)}}$

$\text { Use the following identity: } \cot (x)=\dfrac{\cos (x)}{\sin (x)}$

$=\dfrac{\dfrac{\cos \left(x\right)}{\sin \left(x\right)}}{\csc \left(x\right)}$

$\text { Use the following identity: } \csc (x)=\dfrac{1}{\sin (x)}$

$\sf{=\dfrac{\dfrac{\cos \left(x\right)}{\sin \left(x\right)}}{\dfrac{1}{\sin \left(x\right)}}}$

______________________

Solve $\sf{\dfrac{\dfrac{\cos \left(x\right)}{\sin \left(x\right)}}{\dfrac{1}{\sin \left(x\right)}}}$ :

$\sf{\dfrac{\dfrac{\cos \left(x\right)}{\sin \left(x\right)}}{\dfrac{1}{\sin \left(x\right)}}}$

$=\sf{\dfrac{\cos \left(x\right)\sin \left(x\right)}{\sin \left(x\right)}}$

$\mathrm{Cancel\:the\:common\:factor:}\:\sin \left(x\right)$

$\huge\boxed{\sf{=\cos \left(x\right)}}$

$\mathrm{We\:showed\:that\:the\:two\:sides\:could\:take\:the\:same\:form}$

$\Rightarrow \mathrm{True}$

L.H.S = R.H.S

Hence Proved !!!!