Question

cot^2x/csc^2x + tan^2x/sec^2x =1

identities

Answer 1

\[\cot^2x*\frac{1}{\csc^2x}+\tan^2x*\frac{1}{\sec^2x}\]
=\[\cot^2x*\sin^2x+\tan^2x*\cos^2x\]
=\[\frac{\cos^2x}{\sin^2x}\sin^2x+\frac{\sin^2x}{\cos^2x}{\cos^2x}\]

Answer 2

\[=\cos^2x+\sin^2x=1\]

Answer 3

cot^2x/csc^2x + tan^2x/sec^2x =1
\[(\cot \div \csc)^{2}x+(\tan/\sec)^{2}x\]
\[[(1/\tan)\div (1/\sin)]^{2}x+(\tan \div 1/\cos)^{2}x\]
\[(\sin/\tan)^{2}x+(\cos/\tan)^{2}x\]
sin/tan=[sin/(sin/cos)]^2x=(cos)^2x
tan^2x/sec^2x =(sin/cos)/(1/cos)=sin^2x

Answer 4

\[\cot ^{2}x/\csc ^{2}x = (cosx/sinx) ^{2}/(1/sinx) ^{2} = cos ^{2}x\] & \[\tan ^{2}x/\sec ^{2}x = (sinx/cosx) ^{2}/(1/cosx) ^{2} = sin
^{2}x\]
Since \[cos ^{2}x + sin ^{2}x = 1\], answer is 1

Answer 5

what is the domain restriction?