Question

i will make you brainliest if you will answer tan^a/tan^a-1+ cos^a/sin^a-cos^a=1/1-2cos^a plz answer

Answer 1

the formulae involved are
$({ \tan( a) })^{2} = ({ \sec(a) })^{2} - 1 \\ ({ \sec(a) })^{2} = 1 \div ({ \cos(a) })^{2} \\ ({ \sin(a ) })^{2} + ({ \cos(a) })^{2} = 1 \\ ({ \cos(a) })^{2} - ({ \sin(a) })^{2} = \cos(2a ) \\ \cos(2a) = 2 ({ \cos(a) })^{2} - 1$
=(tan^2a/tan^2a -1)+cos^2a/(sin^a-cos^2a)
=[tan^2a/(sec^2a-1-1)]+[cos^2a/-(cos^2a-sin^2a)]
=[tan^2a/(sec^2a-2)]+cos^2a/-cos2a
=[tan^2a/(sec^2a-2)]+cos^2a/-(2cos^2a-1)
=[tan^2a/{(1/cos^2a)-2}]+[cos^2a/(1-2cos^2a)]
=[tan^2a/{(1-2cos^2a)/cos^2a}]+[cos^2a/(1-2cos^2a)]
=[(tan^2a.cos^2a)/(1-2cos^2a)]+[cos^2a/(1-2cos^2a)]
=[1/(1-2cos^2a)][tan^2a.cos^2a+cos^2a]
=[1/(1-2cos^2a)][(sin^2a/cos^2a)+cos^2a]
=[1/(1-2cos^2a)][sin^2a+cos^2a]
=[1/(1-2cos^2a)](1)
=1/(1-2cos^2a)