Question

# Prove that$$2\sec^{2}\theta -\sec^{4}\theta -2\csc^{2}\theta +\csc^{4}\theta =\cot^{4}\theta -\tan^{4}\theta$$

Solution

## $$\begin{array}{l} 2{ \sec ^{ 2 } }\theta -{ \sec ^{ 4 } }\theta -2{ \csc ^{ 2 } }\theta +{ \csc^{ 4 } }\theta ={ \cot ^{ 4 } }\theta -{ \tan ^{ 4 } }\theta \\ 2\left( { 1+{ { \tan }^{ 2 } }\theta } \right) -{ \left( { 1+{ { \tan }^{ 2 } }\theta } \right) ^{ 2 } }-2\left( { 1+{ { \cot }^{ 2 } }\theta } \right) +{ \left( { 1+{ { \cot }^{ 2 } }\theta } \right) ^{ 2 } } \\ =2+2{ \tan ^{ 2 } }\theta -\left( { 1+{ { \tan }^{ 4 } }\theta +2{ { \tan }^{ 2 } }\theta } \right) -\left( { 2+2{ { \cot }^{ 2 } }\theta } \right) +1+{ \cot^{ 4 } }\theta +2{ \cot ^{ 2 } }\theta \\ =2+2{ \tan ^{ 2 } }\theta -1-{ \tan ^{ 4 } }\theta -2\tan^2{\theta}-2-2{ \cot ^{ 2 } }\theta +1+{ \cot ^{ 4 } }\theta +2{ \cot ^{ 2 } }\theta \\ ={ \cot ^{ 4 } }\theta -{ \tan ^{ 4 } }\theta \,\\ \, \end{array}$$R.H.SHence, proved.Maths

Suggest Corrections

0

Similar questions
View More

People also searched for
View More