Trigonometry! #99

Geometry Level 2

For all 0 < θ < 9 0 ∘ , 0 < \theta < 90^\circ,

sec ⁡ 2 θ + csc ⁡ 2 θ = ? \sqrt{\sec^{2}\theta+\csc^{2}\theta}=\, ?

tan ⁡ θ + cot ⁡ θ \tan\theta+\cot\theta sec ⁡ 2 θ csc ⁡ 2 θ \frac{\sec^{2}\theta}{\csc^{2}\theta} sec ⁡ θ − tan ⁡ θ \sec\theta-\tan\theta sec ⁡ θ + csc ⁡ θ \sec\theta+\csc\theta

Omkar Kulkarni by Omkar Kulkarni , 22, India

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2 solutions

Omkar Kulkarni
Feb 12, 2015

sec ⁡ 2 θ + csc ⁡ 2 θ \sqrt{\sec^{2}\theta+\csc^{2}\theta}

= ( 1 + tan ⁡ 2 θ ) + ( 1 + cot ⁡ 2 θ ) =\sqrt{(1+\tan^{2}\theta)+(1+\cot^{2}\theta)}

= tan ⁡ 2 θ + 2 + cot ⁡ 2 θ =\sqrt{\tan^{2}\theta+2+\cot^{2}\theta}

= tan ⁡ 2 θ + 2 tan ⁡ θ cot ⁡ θ + cot ⁡ 2 θ =\sqrt{\tan^{2}\theta+2\tan\theta\cot\theta+\cot^{2}\theta}

= ( tan ⁡ θ + cot ⁡ θ ) 2 =\sqrt{(\tan\theta+\cot\theta)^{2}}

tan ⁡ θ + cot ⁡ θ \boxed{\tan\theta+\cot\theta}

14 Helpful 0 Interesting 2 Brilliant 2 Confused

How 2 became 2tan cot

Pragati Mishra - 4 years ago

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If you convert tan^2 and cot^2 to sin^2/cos^2 and cos^2/sin^2 you can multiply (sin^2/cos^2)(sin^2/sin^2), (cos^2/sin^2)(cos^2/cos^2) and then 2(cos^2)(sin^2)/cos^2sin^2 and you get = (sin^4+2sin^2cos^2+cos^4)/sin^2cos^2. if you use algebra (x^4+2xy+y^4)= (x^2+y^2)(x^2+y^2) or sin^4...= (sin^2+cos^2)(sin^2+cos^2)/sin^2cos^2 and using trig identities we know that (sin^2+cos^2)=1, so your new answer is 1/sin^2cos^2. sqrt 1/sin^2cos^2= 1/sincos. then we turn the 1 back into sin^2+cos^2 and we get (sin^2/sincos)+(cos^2/sincos) or sin/cos+cos/sin which again converts to tan+cot.

Nicholas Fiacco - 3 years, 3 months ago

2 = 2(1)

tan(x)cot(x) = 1

2 = 2tan(x)cot(x)

Dominik Zmelik - 2 years, 5 months ago

Cot(x). Tan(x) = 1

Muhammad Hatta Hakim - 2 years, 5 months ago
Sandeep Rathod
Feb 9, 2015

1 cos ⁡ 2 x + 1 sin ⁡ 2 x \sqrt{ \dfrac{1}{\cos^2 x} + \dfrac{1}{\sin^2 x}}

cos ⁡ 2 x + sin ⁡ 2 x sin ⁡ 2 x × cos ⁡ 2 x \sqrt{ \dfrac{ \cos^2 x + \sin^2 x}{\sin^2 x \times \cos^2 x}}

1 sin ⁡ x cos ⁡ x \dfrac{1}{\sin x \cos x}

cos ⁡ 2 x + sin ⁡ 2 x sin ⁡ x cos ⁡ x \dfrac{ \cos^2 x + \sin^2 x}{\sin x \cos x}

= tan ⁡ x + cot ⁡ x = \tan x + \cot x

6 Helpful 0 Interesting 1 Brilliant 1 Confused

Okay, thanks!

Omkar Kulkarni - 6 years, 4 months ago

you confuse me

Shiv Mehta - 1 year, 5 months ago

nevermind i understood

Shiv Mehta - 1 year, 5 months ago

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