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Question

If tanθ=sin(α)-cos(α)sin(α)+cos(α), then sin(α)+cos(α) and sin(α)-cos(α) must be equal to


A

2cos(θ),2sin(θ)

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B

2sin(θ),2cos(θ)

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C

2sin(θ),2sin(θ)

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D

2cos(θ),2cos(θ)

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Solution

The correct option is A

2cos(θ),2sin(θ)


Explanation for the correct option:

Step 1. Simplify the equation

tanθ=sin(α)-cos(α)sin(α)+cos(α)

dividing cos(α) to both numerator and denominator

tanθ=sin(α)cos(α)-cos(α)cos(α)sin(α)cos(α)+cos(α)cos(α)tanθ=tan(α)-1tan(α)+1[tanx=sinxcosx]tanθ=tan(α)-tanπ41+tan(α)×tanπ4[tanπ4=1]tanθ=tanα-π4[tan(x-y)=tanx-y1+xy]

Comparing both sides

θ=α-π4α=θ+π4

Step 2. Find the value of sin(α)-cos(α)

sin(α)-cos(α)=sinθ+π4-cosθ+π4sin(α)-cos(α)=cos(θ)sinπ4+sin(θ)cosπ4-cos(θ)cosπ4-sin(θ)sinπ4[sin(x+y)=sin(x)cos(y)+cos(y)sin(x),cos(x+y)=cos(x)cos(y)-sin(x)sin(y)]sin(α)-cos(α)=cos(θ)12+sin(θ)12-cos(θ)12-sin(θ)12sin(α)-cos(α)=12cos(θ)+sin(θ)-cos(θ)+sin(θ)sin(α)-cos(α)=122sin(θ)sin(α)-cos(α)=2sin(θ)

Step 3. Find the value of sin(α)+cos(α)

sin(α)+cos(α)=sinθ+π4+cosθ+π4sin(α)+cos(α)=cos(θ)sinπ4+sin(θ)cosπ4+cos(θ)cosπ4-sin(θ)sinπ4[sin(x+y)=sin(x)cos(y)+cos(y)sin(x),cos(x+y)=cos(x)cos(y)-sin(x)sin(y)]sin(α)+cos(α)=cos(θ)12+sin(θ)12+cos(θ)12-sin(θ)12sin(α)+cos(α)=12cos(θ)+sin(θ)+cos(θ)-sin(θ)sin(α)+cos(α)=122cos(θ)sin(α)+cos(α)=2cos(θ)Hence, option A is correct.


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