Question

Sin(-110°)+tan(290°)/cot(200°)+cos(340°)

Answer 1

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

Given expression is

$\rm \: \dfrac{sin( - 110\degree ) + tan290\degree }{cot200\degree + cos340\degree }$

Let we evaluate each term in to angle between 0° and 90°.

Consider,

$\rm \: sin( - 110\degree )$

We know,

$\boxed{\tt{ sin( - x) \: = \: - \: sinx \: }}$

So, using this, we get

$\rm \: = \: - \: sin110\degree$

$\rm \: = \: - \: sin(90\degree + 20\degree )$

We know,

$\boxed{\tt{ sin(90\degree + x) = cosx \: }}$

So, using this, we get

$\rm \: = \: - \: cos20\degree$

So,

$\rm\implies \:\boxed{\tt{ \: sin( - 110\degree ) = - cos20\degree \: }}$

Now, Consider

$\rm \: tan290\degree$

$\rm \: = \: tan(270\degree + 20\degree )$

We know,

$\boxed{\tt{ \: tan(270\degree + x) = - cotx \: }}$

So, using this identity, we get

$\rm \: = \: - \: cot20\degree$

So,

$\rm\implies \:\boxed{\tt{ tan290\degree \: = \: - \: cot20\degree \: }}$

Now, Consider

$\rm \: cot200\degree$

$\rm \: = \: cot(180\degree + 20\degree )$

We know,

$\boxed{\tt{ cot(180\degree + x) = cotx \: }}$

So, using this result, we get

$\rm \: = \: cot20\degree$

So,

$\rm\implies \:\boxed{\tt{ \: cot200\degree \: = \: cot20\degree \: }}$

Now, Consider

$\rm \: cos340\degree$

$\rm \: = \: cos(360\degree - 20\degree )$

We know,

$\boxed{\tt{ \: cos(360\degree - x) = cosx \: }}$

So, using this, we get

$\rm \: = \: cos20\degree$

So,

$\rm\implies \:\boxed{\tt{ \: cos340\degree \: = \: cos20\degree \: }}$

So, Now, Consider the given expression,

$\rm \: \dfrac{sin( - 110\degree ) + tan290\degree }{cot200\degree + cos340\degree }$

$\rm \: = \: \dfrac{ - cos20\degree - cot20\degree }{cot20\degree + cos20\degree }$

$\rm \: = \: - \: \dfrac{cos20\degree + cot20\degree }{cot20\degree + cos20\degree }$

$\rm \: = \: - \: \dfrac{cot20\degree + cos20\degree }{cot20\degree + cos20\degree }$

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

Hence,

$\rm\implies \: \: \boxed{\tt{ \: \dfrac{sin( - 110\degree ) + tan290\degree }{cot200\degree + cos340\degree } = - \: 1 \: \: }}$

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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

sin (360°-θ)  =  - sin θ

cos (360°-θ)  = cos  θ

tan (360°-θ)  =  - tan θ

csc (360°-θ)  = - csc θ

sec (360°-θ)  =  sec θ

cot (360°-θ)  =  - cot θ