Question

prove that Sinalph
a+cosalpha=√1+sin2alpha​

Answer 1

Answer:

Step-by-step explanation:

2cotα+1sin2α−−−−−−−−−−−−−√

=2cotα+cosec2α−−−−−−−−−−−−−−√

=2cotα+cot2α+1−−−−−−−−−−−−−−−−√

=(1+cotα)2−−−−−−−−−−√

=|1+cotα|

As 3π4<α<π, cotα will be negative.

∴|1+cotα|=−1−cotα

So, optopon -(b) is the correct option.

Answer 2

Step-by-step explanation:

to prove - sin(alpha) + cos(alpha) = under root of 1 + 2sin(alpha)cos(alpha)

since,

sin(alpha) + cos(alpha) = under root of 1 + 2sin(alpha)cos(alpha)

squaring both the sides, we get

[sin(alpha) + cos(alpha) ]² = sin²aplha + cos²aplha + 2sin(alpha)cos(alpha)

[sin(alpha) + cos(alpha) ]² = 1 + 2sin(alpha)cos(alpha)

since ,

sin²aplha + cos²alpha = 1

therefore, shifting the square,

sin(alpha) + cos(alpha) = under root of [1 + 2sin(alpha)cos(alpha)]

hence, proved.

hope its helpful. :)