Question

[0 2 sin 54°• C05 66 (2 sinh.cosB )
(2) 2 cos 50 sino.
(2605A SinB)
(3) 2 sing'. Sin 66
(25inA.sinB)
1) 200578' sin 42 (2LOSA:Sons)
1) 2 C05 54" singo' (2 COSA:sin 13 )​

Answer 1

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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

Answer:

Answer:

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