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Question

sec50°+tan50° is equal to


A

tan(20°)+tan(50°)

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B

2tan(20°)+tan(50°)

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C

tan(20°)+2tan(50°)

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D

2tan(20°)+2tan(50°)

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Solution

The correct option is C

tan(20°)+2tan(50°)


Explanation for the correct option:

Simplify the given expression.

An expression sec50°+tan50° is given.

Simplify the given expression as follows:

sec50°+tan50°=1cos(50°)+tan(50°)sec50°+tan50°=1cos(50°)·cos(20°)cos(20°)+tan(50°)sec50°+tan50°=sin(70°)cos(50°)·cos(20°)+tan(50°)sec50°+tan50°=sin(50°+20°)cos(50°)·cos(20°)+tan(50°)sec50°+tan50°=sin(50°)·cos(20°)+cos(50°)·sin(20°)cos(50°)·cos(20°)+tan(50°)sec50°+tan50°=sin(50°)·cos(20°)cos(50°)·cos(20°)+cos(50°)·sin(20°)cos(50°)·cos(20°)+tan(50°)sec50°+tan50°=sin(50°)cos(50°)+sin(20°)cos(20°)+tan(50°)sec50°+tan50°=tan(50°)+tan(20°)+tan(50°)sec50°+tan50°=tan(20°)+2tan(50°){Since, secθ=1cosθ,sin(A+B)=sin(A)cos(B)+cos(A)sin(B)}

Therefore, the given expression sec50°+tan50° is equal to tan20°+2tan50°.

Hence, the option C is the correct .


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