Question

find value of 4cos20-√3cot20​

Answer 1

Answer:

Here is your answer mate

Step-by-step explanation:

L.H.S=

sin20

∘

1

(4cos20

∘

sin20

∘

−

3

cos20

∘

)

L.H.S=

sin20

∘

2

(sin40

∘

−

2

3

cos20

∘

)

L.H.S=

sin20

∘

1

(2sin40

∘

−2sin60

∘

cos20

∘

)

L.H.S=

sin20

∘

1

[(2sin40

∘

−(sin80

∘

+sin40

∘

)]

L.H.S=

sin20

∘

1

[sin40

∘

−sin80

∘

]

L.H.S=

sin20

∘

1

[2sin(−20

∘

)cos60

∘

]=−1

Answer 2

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

• $\sf{4 cos20^\circ-\sqrt{3}cot20^\circ }$

$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

• Value of $\sf{4 cos20^\circ-\sqrt{3}cot20^\circ }$

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

Given that,

$\sf{4 cos20^\circ-\sqrt{3}cot20^\circ }$

$\sf{= 4cos20^\circ-\sqrt{3}\dfrac{cos20^\circ}{sin20^\circ} }\\\\\sf{= \dfrac{4cos20^\circ.sin20^\circ-\sqrt{3}cos20^\circ }{sin20^\circ} }\\\\\sf{= \dfrac{2sin40^\circ-2.\frac{\sqrt{3}}{2}cos20^\circ }{sin20^\circ} }\\\\\sf{= \dfrac{2sin40^\circ-2.sin60^\circ.cos20^\circ}{sin20^\circ} }\\\\\sf{= \dfrac{2sin40^\circ-(sin80^\circ+sin40^\circ)}{sin20^\circ} }\\\\\sf{= \dfrac{sin40^\circ-sin80^\circ}{sin20^\circ} }\\\\\sf{= \dfrac{2cos60^\circ.sin(-20^\circ)}{sin20^\circ} }\\\\\sf{=-1}$

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