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What is the value of \( \sin 20^{\circ}(\tan 10^{\circ}+\cot 10^{\circ})\)?

Guest Jul 2, 2020

geno3141 · Answer 1 · 2020-07-03T12:16:50

sin(20^o)[ tan(10^o) + cot(10^o) ]

1) = sin(20^o) · tan(10^o) + sin(20^o) · cot(10^o)

2) = sin(20^o) · sin(10^o) / cos(10^o) + sin(20^o) · cos(10^o) / sin(10^o )

3) = sin(20^o) · sin²(10^o) / [ sin(10^o) · cos(10^o ) ] + sin(20^o) · cos²(10^o) / [ sin(10^o) · cos(10^o ) ]

4) = [ sin(20^o) · sin²(10^o) + sin(20^o) · cos²(10^o) ] / [ sin(10^o) · cos(10^o ) ]

5) = [ sin(20^o) · ( sin²(10^o) + cos²(10^o) ) ] / [ sin(10^o) · cos(10^o ) ]

6) = [ sin(20^o) · 1 ] / [ sin(10^o) · cos(10^o ) ]

7) = [ 2 · sin(10^o) · cos(10^o) ] / [ sin(10^o) · cos(10^o ) ]

8) = 2

1) Distribute

2) Rewrite tan as sin/cos; rewrite cot as cos/sin

3) Write with the common denominator of sin(10^o) · cos(10^o )

4) Write with a single denominator

5) Factor out sin(20^o)

6) Use the fact that sin² + cos² = 1

7) Use the fact that sin(2x) = 2sin(x)cos(x)

8) Cancel