- Trending Categories
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Prove the following trigonometric identities:
\( \sec ^{6} \theta=\tan ^{6} \theta+3 \tan ^{2} \theta \sec ^{2} \theta+1 \)
To do:
We have to prove that \( \sec ^{6} \theta=\tan ^{6} \theta+3 \tan ^{2} \theta \sec ^{2} \theta+1 \).
Solution:
We know that,
$\sec ^{2} \theta-\tan^2 \theta=1$.......(i)
$(a+b)^3=a^3+b^3+3ab(a+b)$.........(ii)
Therefore,
$\sec ^{6} \theta=(\sec ^2 \theta)^3$
$=(1+\tan^2 \theta)^3$ [From (i)]
$=1^3+(\tan^2 \theta)^3+3(1)(\tan^2 \theta)(1+\tan^2 \theta)$ [From (ii)]
$=1+\tan^6 \theta+3\tan^2 \theta(1+\tan^2 \theta)$
$=1+\tan^6 \theta+3\tan^2 \theta\sec^2 \theta$ [From (i)]
$=\tan ^{6} \theta+3 \tan ^{2} \theta \sec ^{2} \theta+1$
Hence proved.       
Advertisements