tan+sin=m & tan-sin=n. show that m^2-n^2=4rootm*n
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tan#+Sin#=m .....(1)
tan#-Sin#=n...... .(2)
multiplication by 1 and 2 ,we get
mn=(tan#+sin#) (tan#-sin#)
=tan^2 # -sin^2#
=sin^2# × tan^2#
so
root MN=sin# tan#
now ,squaring (1) &(2) ,and subtract
so,
m^2 - n^2 = (tan#+sin#)^2 -(tan#-sin#)^2
=4×tan# × sin# [(a+b)^2 - (a-b)^2 = 4ab]
=4 root mn (proved)
tan#-Sin#=n...... .(2)
multiplication by 1 and 2 ,we get
mn=(tan#+sin#) (tan#-sin#)
=tan^2 # -sin^2#
=sin^2# × tan^2#
so
root MN=sin# tan#
now ,squaring (1) &(2) ,and subtract
so,
m^2 - n^2 = (tan#+sin#)^2 -(tan#-sin#)^2
=4×tan# × sin# [(a+b)^2 - (a-b)^2 = 4ab]
=4 root mn (proved)
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