Question

If tanθ+sinθ=m and tanθ-sinθ=n, then the value of m²-n² is equal to

• A. 4 mn
• B. 2√(mn)
• C. 4√(mn)
• D. \(2\sqrt(\frac{m}{n})\)

Answer 1

The Correct Option is C

Given that \( \tan\theta+\sin\theta=m \) and \( \tan\theta-\sin\theta=n \), we need to find the value of \( m^2-n^2 \).

We can use the identity:

\[(a+b)(a-b)=a^2-b^2\]

Here, \( a=\tan\theta \) and \( b=\sin\theta \), so substituting in our expressions:

\[m^2-n^2=(\tan\theta+\sin\theta)^2-(\tan\theta-\sin\theta)^2\]

Using the difference of squares formula, we can simplify this to:

\[m^2-n^2=(\tan\theta+\sin\theta+\tan\theta-\sin\theta)(\tan\theta+\sin\theta-\tan\theta+\sin\theta)\]

This simplifies to:

\[m^2-n^2=(2\tan\theta)(2\sin\theta)=4\tan\theta\sin\theta\]

Now, recall that \(\tan\theta=\frac{\sin\theta}{\cos\theta}\). Therefore, substituting gives:

\[4\tan\theta\sin\theta=4\left(\frac{\sin\theta}{\cos\theta}\right)\sin\theta=4\frac{\sin^2\theta}{\cos\theta}\]

This simplifies to:

\[4\left(\frac{\sin^2\theta}{\cos\theta}\right)=4\sqrt{mn}\]

Thus, the value of \( m^2-n^2 \) is \( \boxed{4\sqrt{mn}} \).