Question

Prove the following
$\sf \cos \bigg( \frac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \frac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \} = 1$
- Required step by step explanation
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Answer 1

Step-by-step explanation:

Consider

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \dfrac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \}cos(23π+x)cos(2π+x)cos(

2

3π

+x)cos(2π+x){cot(

2

3π

−x)+cot(2π+x)}cos(23π+x)cos(2π+x)

We Know,

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) = sinxcos(

2

3π

+x)=sinx

\rm \: {cos \: (2\pi + x) }cos(2π+x)

\rm \: \cot \bigg( \dfrac{3\pi}{2} - x \bigg) \: = \: tanxcot(

2

3π

−x)=tanx

\rm \: cot(2\pi + x) \: = \: cotxcot(2π+x)=cotx

So, on substituting all these values, we get

\rm \: = \: sinx \: cosx \: (tanx \: + \: cotx)=sinxcosx(tanx+cotx)

\rm \: = \: sinx \: cosx \: \bigg(\dfrac{sinx}{cosx} + \dfrac{cosx}{sinx}=sinxcosx(

cosx

sinx

+

sinx

cosx

\rm \: = \: sinx \: cosx \: \bigg(\dfrac{ {sin}^{2}x + {cos}^{2}x}{cosx \: sinx}=sinxcosx(

cosxsinx

sin

2

x+cos

2

x

\rm \: = \: 1=1=1=1

Hence,

\boxed{\tt{ \cos \bigg( \frac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \frac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \} = 1}}

cos(

2

3π

+x)cos(2π+x){cot(

2

3π

−x)+cot(2π+x)}=1

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ADDITIONAL INFORMATION :-

Sign of Trigonometric ratios in Quadrants

sin (90°-θ) = cos θ

cos (90°-θ) = sin θ

tan (90°-θ) = cot θ

csc (90°-θ) = sec θ

sec (90°-θ) = csc θ

cot (90°-θ) = tan θ

sin (90°+θ) = cos θ

cos (90°+θ) = -sin θ

tan (90°+θ) = -cot θ

csc (90°+θ) = sec θ

sec (90°+θ) = -csc θ

cot (90°+θ) = -tan θ

sin (180°-θ) = sin θ

cos (180°-θ) = -cos θ

tan (180°-θ) = -tan θ

csc (180°-θ) = csc θ

sec (180°-θ) = -sec θ

cot (180°-θ) = -cot θ

sin (180°+θ) = -sin θ

cos (180°+θ) = -cos θ

tan (180°+θ) = tan θ

csc (180°+θ) = -csc θ

sec (180°+θ) = -sec θ

cot (180°+θ) = cot θ

sin (270°-θ) = -cos θ

cos (270°-θ) = -sin θ

tan (270°-θ) = cot θ

csc (270°-θ) = -sec θ

sec (270°-θ) = -csc θ

cot (270°-θ) = tan θ

sin (270°+θ) = -cos θ

cos (270°+θ) = sin θ

tan (270°+θ) = -cot θ

csc (270°+θ) = -sec θ

sec (270°+θ) = cos θ

cot (270°+θ) = -tan θ