Question

1 + sine/1-sin
=(seco + tane)2
​

Answer 1

Answer:

(sec° + tan °)2 can be written as

{1/cos °+sin°/cos°}2

now take the LCM first then square it

(1+sin°\cos°)2

Step-by-step explanation:

now,

square them separately

(1+sin°)2/cos°

[1+sin°]2 can also be written as 1^2 +sin°^2

now using Identity a^2-b^2= (a-b)(a+b)

we will apply this formula in this (1+sin°)^2

(1-sin°)(1+sin°)/cos^2° _eq1

* we also know the formula that sin^2°+cos^2°=1

we will apply the formula in eq 1

(1+sin°)(1-sin°)/1-sin°^2

now 1-sin^2 ° se upar 1-sin° cancel ho jayega

1+sin °\1-sin° hoga jo RHS ke equal hoga.

hope you find helpful