# Prove the following expression

The equation $$a \cos 2\theta + b \sin2\theta = c$$ has 2 roots (of $$\theta$$): $$\alpha$$ and $$\beta$$. Prove that $$\tan(\alpha + \beta) = \dfrac ba$$.

Note by Pritthijit Nath
2 years, 1 month ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Proceed as follows:-

Let the eq.n be :→ a(cos2p) + b(sin2p) = c (just for convenience in writing)

Now we substitute,

x = tan(p)

So that,

Sin2p = 2x/(1+x^2) &

Cos2p = (1-x^2)/(1+x^2)

Now our eq.n reduces to→

a [2x/(1+x^2)] + b [(1-x^2)/(1+x^2)] =c

On further simplification u get a quadratic eq. in x as follows→

(b+c)•x^2 - 2ax + c-b = 0

Now let the two roots be tanA & tanB,

Then by vieta's formulae,

TanA + TanB = 2a/(b+c) &

TanA●TanB = (c-b)/(c+b)

Now use the formula of Tan (A+B)→

Tan (A+B) = (TanA + TanB)/(1-TanA●TanB)

To get→

| Tan(A+B) = a/b |★ ( Ans.)

Please feel free to correct and suggest me wherever im wrong ...

Thank you.!

- 2 years, 1 month ago