Question

1+tanx tan x/2= secx = tanx cotx/2 - 1....prove it

Answer 1

solution by biology student. I am from Biology stream

Answer 2

It is proven that 1+tanx tan x/2= secx = tanx cotx/2 - 1  

Given:

1+tanx tan x/2= secx = tanx cotx/2 - 1

To find:

Prove that 1+tanx tan x/2 = secx = tanx cotx/2 - 1

Solution:

Take 1+tanx tan x/2 = secx  

Here LHS =  1+tanx tan x/2

=  $1 + (\frac{sin x}{cos x} )(\frac{sin (x/2)}{cos (x/2)} )$             [ ∵ tan = sin/cos ]

= $\frac{cos x cos (x/2)+ sin x sin (x/2) }{cosx cos(x/2)}$       [ ∵ cosA cosB + sinA sinB = cos (A-B) ]

=  $\frac{cos (x-\frac{x}{2} ) }{cos x cos(x/2)}$  

=  $\frac{cos (\frac{x}{2} ) }{cos x cos(x/2)}$

=  $\frac{1}{cos x }$ = sec x  = RHS

⇒ ∴ LHS = RHS

⇒ 1+tanx tan x/2 = sec x ---(1)  

Take tanx cotx/2 - 1 = secx  

Take LHS tanx cotx/2 - 1  

⇒  $\frac{sinx }{cosx} (\frac{cos (x/2)}{sin(x/2)}) - 1$

=  $\frac{sinx cos (x/2)- cosx sin(x/2)}{cosx sin(x/2)}$          [ ∵ sinA cosB - cosA sinB = sin(A-B) ]

=  $\frac{sin(x - \frac{x}{2} ) }{cosx sin(x/2)}$

=  $\frac{sin(\frac{x}{2} ) }{cosx sin(x/2)}$

=  $\frac{1 }{cosx}$ = sec x = RHS

⇒ LHS = RHS

⇒  tanx cotx/2 - 1 = secx ------ (2)

From (1) and (2)

1+tanx tan x/2= secx = tanx cotx/2 - 1  

Hence,

It is proven that 1+tanx tan x/2= secx = tanx cotx/2 - 1  

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