Doubt in trigo

Use, tan(90-@) = cot@

Expression wud reduce to-

→ tan9+ cot9 -cot27 -tan27
=sec²9/tan9 - sec²27/tan27

Changing to sin, cos-
= 2/sin18 - 2/sin54

Then use the standard values....Ans comes out to be 4.

What if we aren't given the values? Because values of 0,30,45,60 and 90 degrees are known to us.

Then there's none other better option than to leave it at 2/sin18 - 2/sin54.

Waise, i have never been able to remember those values myself, nor have i ever bothered or tried to... :P, they're only 2b looked at when needed.

Because I saw this in an MCQ paper. Four options were given:

1 , 2, 3, 4

How do we know what the answer is, that too at class X level?

Incase you don't remember the value of sin 18° then, calculate it as follows:

Let A = 18°
=> 5A = 90°
=> 3A + 2A = 90°
=> 3A = 90° - 2A
=> Sin 3A = Cos 2A [since , sin (90°-A) = cos A]
=> 3 sinA - 4sin³A = 1 - 2sin²A
=> -4sin³A + 2sin²A + 3 sinA - 1 = 0
=> 4sin³A - 2sin²A - 3 sinA + 1 =0

On putting sin A = x we get,

4x³ - 2x² -3x + 1 = 0

by trial 1 is a root of the above equation.. and so by facgtor theorem (x - 1) will be a factor
of the above eqn...

4x³ - 4x²+ 2x² - 2x - x + 1 = 0

=> 4x² (x - 1) + 2x (x - 1) - 1(x - 1) = 0
=> (x - 1)(4x² + 2x - 1) = 0

On solving you will get the value of x.

you'll get x = √5 - 1/4 or, 1 - √5 / 4
but since, x can't be negative....

so x = √5 - 1 / 4...

or sin 18° = √5 - 1 / 4...