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If tanθ + sinθ = m and tanθ – sinθ = n, then prove that m2 – n2 = 4sinθ tanθ [Hint: m + n = 2tanθ, m – n = 2sinθ, then use m2 – n2 = (m + n)(m – n)]

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Question

If tanθ + sinθ = m and tanθ – sinθ = n, then prove that m2 – n2 = 4sinθ tanθ 
[Hint: m + n = 2tanθ, m – n = 2sinθ, then use m2 – n2 = (m + n)(m – n)]

Theorem
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Solution

tanθ + sinθ = m   ......(i)

tanθ – sinθ = n  ......(ii)

Adding equations i and ii

2tanθ = m + n   ......(iii)

Subtracting equation ii from i

We get,

2sinθ = m – n  ......(iv)

Multiplying equations (iii) and (iv)

2sinθ(2tanθ) = (m + n)(m – n)

⇒ 4sinθ tanθ = m2 – n2

Hence,

m2 – n2 = 4sinθ tanθ

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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Chapter 3: Trigonometric Functions - Exercise [Page 53]

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NCERT Exemplar Mathematics [English] Class 11
Chapter 3 Trigonometric Functions
Exercise | Q 10 | Page 53

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