×

# trignometry challenge

GIVEN (a^2-b^2)sintheta +2abcostheta = a^2 +b^2

TO FIND = tantheta

i am not able to sole this problem need help...

Note by Avn Bha
3 years, 4 months 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:

how did u get it?

- 3 years, 4 months ago

@avn bha , My answer is quite long. So please try to find any shorter way (if any) and do tell me... Dividing the given eqn. by $$cos{ \theta }$$. You get the equation in terms of $$\tan { \theta } \quad and\quad \sec { \theta }$$. Squaring both sides and converting $$\sec ^{ 2 }{ \theta } =1+\tan ^{ 2 }{ \theta }$$ and simplifying, u get an equation, quadratic in $$\tan { \theta }$$ which looks like

$$4{ a }^{ 2 }{ b }^{ 2 }\tan ^{ 2 }{ \theta } -4ab({ a }^{ 2 }-{ b }^{ 2 })\tan { \theta } +{ ({ a }^{ 2 }-{ b }^{ 2 }) }^{ 2 }=0\\ Using\quad quadratic\quad formula\quad you\quad get\quad \\ \tan { \theta } =\frac { 4ab({ a }^{ 2 }-{ b }^{ 2 }) }{ 8{ a }^{ 2 }{ b }^{ 2 } } =\frac { { a }^{ 2 }-{ b }^{ 2 } }{ 2ab }$$

CHEERS!!!:)

- 3 years, 4 months ago

$$\tan \theta = \frac{a^{2}-b^{2}}{2ab}$$

- 3 years, 4 months ago