Prove that ` (1)/(secA+tanA)=secA-tanA`

Answer

C

Explanation

a. `cos 2A +cos 2B + cos 2C =-1 `
`rArr -1 -4 cos A cos B cos C =-1 `
`rArr cos A cos B cos C =0`
`rArr cos A =0 or cos B = 0 or cos C =0`
Thus, triangle is right angled.
b. `tan A gt 0, tan B ge 0 and tan A tan B lt 1`
`tan C = ( tan A + tanB)/(tan A tan B -1) lt 0`
`therefore angle C gt 90^(@)`.
c. `cos^(3) A + cos^(3)B + cos^(3)C = 3 cos A cos B cosC`
`rArr (cos A + cos B + cos C)(cos ^(2) A + cos^(2) B + cos^(2)C - cos A cosB - cos B cosC - cos A cos C) =0`
`rArr cot A + cos B+ cos C =0`
or `cos ^(2) A + cos^(2)B + cos ^(2)C -coA cos B - cos B cos C - cosA cos C =0`
`rArr 1+4 sin (A//2)sin (B//2 ) sin (C//2) =0`
or `cos A = cos B = cos C`
`rArr sin(A//2) sin (B//2) sin (C//2) = -1//4`( which is not possible as `sin(A//2), sin(B//2), sin (C//2) gt 0`)
`or cos A =cos B =cos C `
Thus, triangle is equilateral.
d. `cotA gt 0, cot B gt 0 and cot A cot B lt 1`
`therefore tan A gt 0, tan Bgt 0 and tan A tan B gt 1`
`therefore tan C gt 0`
`therefore angle C lt 90^(@)`