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Write in terms of sine and cosine, Simplify the trig expression to a power of a single trig function?

0 votes

Write in terms of sine and cosine: 
cos^2 θ(tan^2 θ+1) = 

(sec θ-cos θ)/(sin θ) = 
------------------------------------------------ 
Simplify the trig expression to a power of a single trig function: 
(sec^2 x-1)/(sec^2 x) 

 

asked Nov 15, 2014 in TRIGONOMETRY by anonymous

3 Answers

0 votes

1)

cos²θ ( tan²θ + 1 )

Substitute : tanθ = sinθ / cosθ

= cos²θ [ (sinθ / cosθ)² + 1 ]

= cos²θ [ (sin²θ / cos²θ) + 1 ]

Rewrite the expression with common denominator.

= cos²θ [ (sin²θ+1) / cos²θ ]

= [ cos²θ / cos²θ ](sin²θ+1)

= (sin²θ+1)

= 1 + sin²θ

Solution :

cos²θ ( tan²θ + 1) = 1 + sin²θ

answered Nov 15, 2014 by Shalom Scholar
0 votes

2)

(secθ - cosθ) / (sinθ)

Substitute : secθ = 1 / cosθ

= [(1 / cosθ) - cosθ] / (sinθ)

= [(1 - cos²θ) / cosθ ] / (sinθ)

= [(1 - cos²θ) / sinθcosθ ]

Substitute :  sin²θ = 1 - cos²θ

= [(sin²θ) / sinθcosθ ]

= sinθ / cosθ

Solution :

(secθ - cosθ) / (sinθ) = sinθ / cosθ

answered Nov 15, 2014 by Shalom Scholar
0 votes

3)

(sec² x-1)/(sec² x)

Substitute : tan²x = sec² x - 1

= (tan² x)/(sec² x)

Substitute : tanx = sinx/cosx and secx = 1/cosx

= [(sinx/cosx)²]/(1/cosx

= (sin²x/cos²x)/(1/cos²x)

= (sin²x)(cos²x)/cos²x

= (sin²x)

= (sinx)²

Solution :

(sec² x-1)/(sec² x) = (sinx)²

answered Nov 15, 2014 by Shalom Scholar

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