Question

Prove that : 1/. Sec x- tan X - 1/cos X = 1/ cos X - 1/ sec X+ tan x

Answer 1

Answer:

TO PROVE : 1/( sec x - tan x ) - 1/cos x = 1/cos x -  1/( sec x  + tan x )

LHS :

1/( sec x - tan x ) - 1/cos x

= sec x + tan x -sec x

= sec x - ( sec x-tan x )

= 1/cos x - (sec^2 x-tan^2 x) / ( sec x + tan x )

=1/cos x - 1/( sec x + tan x )

= RHS

Hope this helps!

Answer 2

Qᴜᴇsᴛɪᴏɴ :-

Prove that : 1/. Sec x- tan X - 1/cos X = 1/ cos X - 1/ sec X+ tan x

Sᴏʟᴜᴛɪᴏɴ :-

Taking LHS ,

→ 1/(Secx- tanx) - (1/cosx)

Multiplying by (secx + tanx) in the numerator and denominator of First Term and using (1/cosA) = secA in second Term we get,

→ [(secx + tanx)/{(secx +tanx)(secx - tanx)}] - secx

Now using (a + b)(a - b) in Denominator we get,

→ [(secx + tanx)/(sec²x - tan²x)] - secx

Now putting sec²A - tan²A = 1 in Denominator,

→ secx + tanx - secx

→ tanx

Now, Adding and subtracting secx , we get

→ secx + tanx - secx

→ secx + (tanx - secx)

Taking (-1) common & using (1/cosA) = secA

→ (1/cosx) - (secx - tanx)

Now Multiply and Divide second Term by (secx + tanx), we get ,

→ (1/cosx) - [{(secx - tanx)(secx + tanx)} / (secx + tanx)]

→ (1/cosx) - (sec²x - tan²x)/(secx + tanx)

Putting Second term of Numerator = 1 by sec²A - tan²A = 1 Now,

→ (1/ cosx) - 1/(secx + tanx) = RHS (Hence, Proved.)