Trig Identities Abound

Geometry Level 2

Where defined, cot ⁡ θ − cot ⁡ θ ( cos ⁡ θ − sin ⁡ θ ⋅ tan ⁡ θ cos ⁡ θ ) = ? \Large \cot \theta - \cot \theta \left(\frac{\cos \theta - \sin \theta \cdot \tan \theta}{\cos \theta}\right) = \ ?

cos ⁡ θ \cos \theta csc ⁡ θ \csc \theta sin ⁡ θ \sin \theta tan ⁡ θ \tan \theta sec ⁡ θ \sec \theta cot ⁡ θ \cot \theta

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Jade Mijares by Jade Mijares , 30, Philippines

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4 solutions

Michael Fuller
Nov 27, 2015

Using the relevant trigonometric identites ,

cot ⁡ θ − cot ⁡ θ ( cos ⁡ θ − sin ⁡ θ tan ⁡ θ cos ⁡ θ ) \cot \theta - \cot \theta \Big(\frac{\cos \theta - \sin \theta \tan \theta}{\cos \theta}\Big)

= cot ⁡ θ ( 1 − cos ⁡ θ − sin ⁡ θ tan ⁡ θ cos ⁡ θ ) =\cot \theta \Big(1-\frac{\cos \theta-\sin \theta \tan \theta}{\cos \theta}\Big)

= cot ⁡ θ ( sin ⁡ θ tan ⁡ θ cos ⁡ θ ) =\cot \theta \Big(\frac{\sin \theta \tan \theta}{\cos \theta}\Big)

= ( sin ⁡ θ cos ⁡ θ ) =\Big(\frac{\sin \theta}{\cos \theta}\Big)

= tan ⁡ θ =\large \color{#20A900}{\boxed{\tan \theta}}

16 Helpful 0 Interesting 0 Brilliant 0 Confused

i too did in the same way

PSN murthy - 5 years, 6 months ago

Nice! Same solution here too!

Ryoha Mitsuya - 5 years, 6 months ago
Jade Mijares
Nov 26, 2015

cot ⁡ θ − cot ⁡ θ ( cos ⁡ θ − sin ⁡ θ tan ⁡ θ cos ⁡ θ ) \cot \theta - \cot \theta \Big( \frac{\cos \theta - \sin \theta \tan \theta}{\cos \theta} \Big) = cos ⁡ θ sin ⁡ θ − cos ⁡ θ sin ⁡ θ ( cos ⁡ θ − sin ⁡ θ tan ⁡ θ cos ⁡ θ ) =\frac{\cos \theta}{\sin \theta} - \frac{\cos \theta}{\sin \theta} \Big( \frac{\cos \theta - \sin \theta \tan \theta}{\cos \theta} \Big) = cos ⁡ θ sin ⁡ θ − cos ⁡ θ − sin ⁡ θ ( sin ⁡ θ cos ⁡ θ ) sin ⁡ θ =\frac{\cos \theta}{\sin \theta} - \frac{\cos \theta - \sin \theta (\frac{\sin \theta}{\cos \theta})}{\sin \theta} = cos ⁡ θ sin ⁡ θ − ( cos ⁡ 2 θ − sin ⁡ 2 θ cos ⁡ θ ) sin ⁡ θ =\frac{\cos \theta}{\sin \theta} - \frac{(\frac{\cos^2 \theta - \sin^2 \theta}{\cos \theta})}{\sin \theta} = cos ⁡ θ sin ⁡ θ − cos ⁡ 2 θ − sin ⁡ 2 θ cos ⁡ θ sin ⁡ θ =\frac{\cos \theta}{\sin \theta} - \frac{\cos^2 \theta - \sin^2 \theta}{\cos \theta \sin \theta} = cos ⁡ 2 θ − ( cos ⁡ 2 θ − sin ⁡ 2 θ ) cos ⁡ θ sin ⁡ θ =\frac{\cos^2 \theta - (\cos^2 \theta - \sin^2 \theta)}{\cos \theta \sin \theta} = sin ⁡ 2 θ cos ⁡ θ sin ⁡ θ =\frac{\sin^2 \theta}{\cos \theta \sin \theta} = sin ⁡ θ cos ⁡ θ =\frac{\sin \theta}{\cos \theta} = tan ⁡ θ =\boxed{\tan \theta}

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Lu Chee Ket
Dec 8, 2015

C S − C S C − S S C C \displaystyle \frac{C}{S} - \displaystyle \frac{C}{S} \displaystyle \frac{C - S \displaystyle \frac{S}{C}}{C} = C S − 1 S C − S S C 1 \displaystyle \frac{C}{S} - \displaystyle \frac{1}{S} \displaystyle \frac{C - S \displaystyle \frac{S}{C}}{1} = C S − C S + S S C S \displaystyle \frac{C}{S} - \displaystyle \frac{C}{S} + \displaystyle \frac{S \displaystyle \frac{S}{C}}{S} = S C = tan ⁡ θ \displaystyle \frac{S}{C} = \tan \theta

Answer: tan ⁡ θ \boxed{\tan \theta}

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Adnan Khan
Nov 27, 2015

I am agree with jade mijares

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