- 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 that:$ \sin ^{2} A \cos ^{2} B-\cos ^{2} A \sin ^{2} B=\sin ^{2} A-\sin ^{2} B $
To do:
We have to prove that \( \sin ^{2} A \cos ^{2} B-\cos ^{2} A \sin ^{2} B=\sin ^{2} A-\sin ^{2} B \).
Solution:
We know that,
$\sin^2 A+\cos^2 A=1$
$\operatorname{cosec}^2 A-\cot^2 A=1$
$\sec^2 A-\tan^2 A=1$
$\cot A=\frac{\cos A}{\sin A}$
$\tan A=\frac{\sin A}{\cos A}$
$\operatorname{cosec} A=\frac{1}{\sin A}$
$\sec A=\frac{1}{\cos A}$
Therefore,
$\sin ^{2} A \cos ^{2} B-\cos ^{2} A \sin ^{2} B=\sin ^{2} A (1-\sin ^{2} B)-(1-\sin ^{2} A) \sin ^{2} B$
$=\sin ^{2} A -\sin ^{2} A\sin ^{2} B-\sin ^{2} B+\sin ^{2} A \sin ^{2} B$
$=\sin ^{2} A-\sin ^{2} B$
Hence proved.
Advertisements