Question

do this question and get 1000 rupees

Answer 1

Let's start by taking the LCM we'll get
(cos10-√3sin10)/sin10cos10
Let's Divide and Multiply numerator by 2 we'll get
2{1/2cos10-√3/2sin10}/sin10cos10
Now we know that 1/2=cos60 and √3/2=sin60
Hence we'll get
2{cos60cos10-sin60sin10}/sin10cos10
Now we'll Multiply and Divide Denominator by 2 then we'll get
2{cos60cos10-sin60sin10}/{1/2*2sin10cos10}
Since we know
cosAcosB-sinAsinB=cos(A+B)
and
2sinAcosA=sin2A
we'll get
2cos(60+10)/{1/2*sin20}
4cos70/sin20
4cos70/sin(90-70)
4cos70/cos70
4

Hence answer 4 is the correct response.


Now where is my 1000/-.
;)

Answer 2

LHS :

sin10o1−cos10o3

=2[2sin10o1−2cos10o3]

=2[sin10osin30o−cos10ocos30o]

=2[sin10ocos10osin30ocos10o−cos30osin10o]

=2[sin10ocos10osin20o]

=2×2[sin10ocos10osin10ocos10o]

=4 = RHS