Trig-algebra

Algebra Level 3

tan ⁡ θ + sin ⁡ θ = m \tan \theta + \sin \theta = m and m 2 − n 2 = 4 m n m^{2} - n^{2} = 4\sqrt{mn}

What is n?

PS: Would love to see how you solved it!

sin ⁡ θ \sin \theta tan ⁡ θ \tan \theta tan ⁡ θ − sin ⁡ θ \tan \theta - \sin \theta

Shantanu Rahman by Shantanu Rahman , 20, Bangladesh

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1 solution

Krutarth Patel
Nov 16, 2015

Let t a n θ − sin ⁡ θ tan \theta - \sin \theta = n m n = ( t a n θ + s i n θ ) ( t a n θ − s i n θ ) = t a n 2 θ − s i n 2 θ = s i n 2 θ ( s e c 2 θ − 1 ) = s i n 2 θ t a n 2 θ \begin{aligned} mn & = (tan \theta + sin \theta)(tan \theta - sin \theta) \\ & = tan^{2} \theta - sin ^ {2} \theta \\ & = sin ^ {2} \theta (sec ^ {2} \theta - 1) \\ & = sin ^ {2} \theta tan ^ {2} \theta \end{aligned} . Now, ( m + n ) 2 − ( m − n ) 2 = ( t a n θ + s i n θ ) 2 − ( t a n θ − s i n θ ) 2 = 4 t a n θ s i n θ = 4 t a n 2 θ s i n 2 θ = 4 m n (m + n) ^{2} - (m - n)^{2} = (tan \theta + sin \theta)^{2} - (tan \theta - sin \theta)^{2} = 4 tan \theta sin \theta = 4\sqrt{tan ^{2} \theta sin ^{2} \theta} = 4\sqrt{mn} Hence t a n θ − s i n θ \boxed{tan \theta - sin \theta} is the answer.

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Can you do it without assuming the answer?

Shantanu Rahman - 5 years, 7 months ago

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