Question

tan s+ sec s -1 ÷tans - sec s + 1 = 1+ sin s ÷ cos s​

Answer 1

Answer:

$\frac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}=\frac{1+sin\theta}{cos\theta}$

Step-by-step explanation:

Formula used:

$sec^2\theta-tan^2\theta=1$

Now,

$\frac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}$

$=\frac{tan\theta+sec\theta-(sec^2\theta-tan^2\theta)}{tan\theta-sec\theta+1}$

$=\frac{(sec\theta+tan\theta)-(sec\theta-tan\theta)(sec\theta+tan\theta)}{tan\theta-sec\theta+1}$

$=(sec\theta+tan\theta)(\frac{1-(sec\theta-tan\theta)}{tan\theta-sec\theta+1})$

$=(sec\theta+tan\theta)(\frac{1-sec\theta+tan\theta}{tan\theta-sec\theta+1})$

$=(sec\theta+tan\theta)(\frac{tan\theta-sec\theta+1}{tan\theta-sec\theta+1})$

$=sec\theta+tan\theta$

$=\frac{1}{cos\theta}+\frac{sin\theta}{cos\theta}$

$=\frac{1+sin\theta}{cos\theta}$

Answer 2

Answer:

Proved

(Tan s + Sec s - 1)/(Tan s - Sec s + 1)  = (1 + Sin s)/Cos s

Step-by-step explanation:

Question is

(Tan s + Sec s - 1)/(Tan s - Sec s + 1)  = (1 + Sin s)/Cos s

LHS

= (Tan s + Sec s - 1)/(Tan s - Sec s + 1)

multiplying numerator & Denominator by Cos s

= (Sin s + 1 - Cos s)/(Sin s - 1 + Cos s)

= ( (Sin s - Cos s) + 1) / ( (Sin s + Cos s) - 1)

now Dividing & Multiplying with (Sin s + Cos s ) + 1

= (Sin² s - Cos² s + Sin s - Cos s + Sin s + cos s + 1) / (Sin² s + Cos² s + 2Sin s Cos s - 1)

Using Sin² s + Cos² s = 1

= ( Sin² s + 2Sin s + (1 - Cos² s) ) /( 1 + 2 Sin s Cos s  - 1)

=  ( Sin² s + 2Sin s + Sin² s ) /(2 Sin s Cos s)

= ( 2Sin² s + 2Sin s) /(2 Sin s Cos s)

Cancelling 2 Sin s from numerator Denominator

= (Sin s + 1 )/ Cos s

= (1 + Sin s)/ Cos s

= RHS

QED