Question

Prove that : sec A + tan A/sec A - tan A = (1+ sinA/cos A)^2.​

Answer 1

Answer:

given : Sec A + tan A =( 1+ sin A)^2

sec A - tan A. ( cos A )

LHS:

= Sec A + tan A / sec A - tan A

= (1/ cos A + sin A / cos A) / ( 1/cos A - sin A / cos A)

= (cos A / 1 + sin A) ( 1- sin A / cos A)

= (1 + Sin A ) / 1 - sin A)

multiplying by ( 1+ sin A ) / ( 1- sin A)

= ( 1+ sin ^ 2 A) / ( 1- sin ^ 2 A)

= ( 1+ sin ^ 2 A) / (cos A )

HENCE PROVED

HOPE THIS ANSWER HELPS YOU........

Answer 2

Step-by-step explanation:

$\huge \underline \mathbb {SOLUTION:-}$

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0≤r<6.

Now substituting the value of r, we get,

If r = 0, then a = 6q

Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.

If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2. A positive integer can be either even or odd Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.