Trigonometric Ratios and Identities

Given,"‌t2−9t+8=0,‌ is satisfied by ‌
‌ Now, ‌alocos2x+cos4x+cos6x+...∞)‌logc‌2=blogca
∴‌‌e(cos2x+cos4x+...∞)‌log‌2
=2(cos2x+cos4x+∞)logee=2cos2x+cos4x+......∞
‌ Here, ‌cos2x+cos4x+...∞‌ are in GP, ‌
‌ where ‌a=cos2x
∴‌‌r=cos2x<1
∴‌‌S∞=‌

a

1−r

∴‌‌S∞=‌

cos2x

1−cos2x

=‌

cos2x

sin2x

=cot2x
∴‌‌cos2x+cos4x+...∞=cot2x
‌ Now, ‌2cos2x+cos4x+...∞=2cot2x
‌ Now, roots of equation ‌t2−9t+8=0,‌ are ‌
‌‌‌(t−1)(t−8)=0
‌‌‌‌‌t=1,8
⇒‌‌2cot2x=1‌ or ‌8
⇒‌‌cot2x=0‌ or ‌cot2x=3⇒cot‌x=0‌ or ‌cot‌x=√3
\But here, 0<x<‌

π

2

.
∴cot‌x=0 not possible.
Hence, cot‌x=√3 is only possible value.
Now, ‌

2‌sin‌x

sin‌x+√3‌cos‌x

Dividing numerator and denominator by sin‌x, we get
‌‌

2

1+√3‌cot‌x

=‌‌‌

2

1+√3×√3

=‌‌‌

2

1+3

=‌

2

4

=‌

1

2

[∵cot‌x=√3]