Question

If tan A = n tanB and sinA = m sinB, then prove that cos^(2) = (m^2 - 1)/(n^2 - 1)

Answer 1

Given:

• tan A = n tanB

• sinA = m sinB

To be prove:

$\boxed{cos^{2}= \frac{m^{2}-1}{n^{2}-1}}$

We have to find the $cos^{2}A$ in terms of m and n.

So, we have to eliminate ∠B from the given relations.

Now,

$tan A=n\ tan B \implies tan B = \frac{1}{n} tan A$

$\implies cot B= \frac{n}{tan A}\ \ \ \ [\therefore cotB = \frac{1}{tanB}]$

And also,

$sinA= m\ sinB$

$\implies sinB = \frac{1}{m} sinA$

$\implies cosecB = \frac{m}{sinA}\ \ \ \ [\therefore sinB = \frac{1}{cosecB}]$

We know that

$\boxed{cosec^{2}B - cot^{2}B = 1}$

So,

Substituting the values of $cotB$ and $cosec B$, we will get,

$\boxed{(\frac{m}{sinA})^{2} - (\frac{n}{tanA})^{2} =1}$

$\implies \frac{m^{2}}{sin^{2}A} - \frac{n^{2}}{tan^{2}A} =1$

$\implies \frac{m^{2}}{sin^{2}A} - \frac{n^{2}}{(\frac{sin^{2}A}{cos^{2}A})} =1\ \ \ \ [\therefore tanA = \frac{sinA}{cosA}]$

$\implies \frac{m^{2}}{sin^{2}A} -\frac{n^{2}cos^{2}A}{sin^{2}A}=1$

$\implies \frac{m^{2}-n^{2}cos^{2}A}{sin^{2}A}=1$

$\implies m^{2} - n^{2}cos^{2}A=sin^{2}A$

⇒ m² - n² cos²A = 1 - cos²A    [∴ sin²A = 1 - cos²A]

⇒ m² - 1 = n² cos²A - cos²A

⇒ m² - 1 = cos²A (n² - 1)

$\therefore \boxed{cos^{2}A = \frac{m^{2}-1}{n^{2}-1}}$

Hence proved.

Trigonometric identities -

• sin²Ф + cos²Ф = 1
• sec²Ф = 1 + tan²Ф
• cosec²Ф = cot²Ф + 1