If tanθ = 15/8 and cosθ is negative, find the value of:

[sin(-θ)-cosθ] / [tan(-θ)+sec(-θ)]

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Question

If tanθ = 15/8 and cosθ is negative,
find the value of:
[sin(-θ)-cosθ] /
[tan(-θ)+sec(-θ)]

Santosh Kumar · Accepted Answer

$Given, \tan θ = \frac{15}{8}$

AB = 15, BC = 8

$AC = \sqrt{{AB}^{2} + {BC}^{2}} \Rightarrow AC = \sqrt{{(15)}^{2} + (8)^{2}} \Rightarrow AC = \sqrt{225 + 64} \Rightarrow AC = 17$

Since, tan θ is (+)ve and tan is always (+)ve in the III quadrant, so value of θ lies in the third quadrant.

$\sin θ = \frac{AB}{AC} = \frac{- 15}{17} cos θ = \frac{BC}{AC} = \frac{- 8}{17} sec θ = \frac{1}{\cos θ} = \frac{- 17}{8}$

(In 3rd quadrant, value of sin θ, cos θ and sec θ is (–)ve.)

Now,

$\frac{[\sin (- θ) - \cos θ]}{[\tan (- θ) + \sec (- θ)]} = \frac{- \sin θ - \cos θ}{- \tan θ + \sec θ} = \frac{- (\frac{- 15}{17}) - (\frac{- 8}{17})}{\frac{- 15}{8} - \frac{17}{8}} = \frac{(\frac{15}{17} + \frac{8}{17})}{- 4} = \frac{- 23}{68}$

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If tanθ = 15/8 and cosθ is negative, find the value of: [sin(-θ)-cosθ] / [tan(-θ)+sec(-θ)]

If tanθ = 15/8 and cosθ is negative, find the value of:

[sin(-θ)-cosθ] / [tan(-θ)+sec(-θ)]