Question

Evaluate:2sin68/cos22 - 2cot15/5tan75 - 3tan45 tan20 tan40 tan50 tan70/5

Answer 1

The Formulas to be used in this Questions are:=>cos(90 - x) = sin x   e.g. cos 22 = cos(90-68)= sin 22
=>tan(90 - x) = cot x    e.g. cot 15 = cot(90-75) = tan 15
=>tanx.cotx = 1

Now Question is,
2sin68/cos22 - 2cot15/5tan75 - (3/5)tan45 tan20 tan40 tan50 tan70
=2sin68/sin68 - 2tan75/5tan75 - (3/5)1 tan20 tan40 cot40 cot 20

=2 - 2/5 - 3/5= 1
This is the Answer

Answer 2

Before solving, the following formulas must be kept in mind.
sin x = cos (90-x)
tan x = cot (90-x)
tan(x)×cot(x) = 1
sin 68° = cos 22°
cot 15° = tan 75°
tan 45° = 1
tan 50° = cot 40° & tan 70° = cot 20°
The problem then reduces to 2 sin 68°/sin 68° - 2 cot 15°/5 cot 15° - 3 tan 20° tan 40° cot 40° tan 40°/5 = 2-2/5-3/5 = 2-1 = 1.