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
- Spam will be reported on spot.​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider

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

We know,

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

$\rm \: cos \: (2\pi + x) = cosx$

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

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

So, on substituting all these values, we get

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

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

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

$\rm \: = \: 1$

Hence,

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

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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

Answer 2

$\large\underline{\sf{Solution-}}$

Consider

$\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \dfrac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \}cos( </p><p>2</p><p>3π</p><p></p><p> +x)cos(2π+x){cot( </p><p>2</p><p>3π</p><p></p><p> −x)+cot(2π+x)}</p><p>$

We know,

$\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) = sinxcos( </p><p>2</p><p>3π</p><p></p><p> +x)=sinx</p><p></p><p>\rm \: cos \: (2\pi + x) = cosxcos(2π+x)=cosx</p><p></p><p>\rm \: \cot \bigg( \dfrac{3\pi}{2} - x \bigg) \: = \: tanxcot( </p><p>2</p><p>3π</p><p></p><p> −x)=tanx$

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

So, on substituting all these values, we get

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

$\rm \: = \: sinx \: cosx \: \bigg(\dfrac{sinx}{cosx} + \dfrac{cosx}{sinx} \bigg)=sinxcosx( </p><p>cosx</p><p>sinx</p><p></p><p> + </p><p>sinx</p><p>cosx</p><p></p><p> )$

$\rm \: = \: sinx \: cosx \: \bigg(\dfrac{ {sin}^{2}x + {cos}^{2}x}{cosx \: sinx} \bigg)=sinxcosx( </p><p>cosxsinx</p><p>sin </p><p>2</p><p> x+cos </p><p>2</p><p> x</p><p></p><p> )$

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

Hence,

$\boxed{\tt{ \cos \bigg( \frac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \frac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \} = 1}} </p><p>cos( </p><p>2</p><p>3π</p><p></p><p> +x)cos(2π+x){cot( </p><p>2</p><p>3π</p><p></p><p> −x)+cot(2π+x)}=1</p><p>$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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