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The expression cos2θ3−4sin2θ+2tan2θ can be simplified as:
0
3tan2θ
3+tan2θ
3
3+tan2θ
Substitute θ=0°. The expression becomes 13−0+0=3. Checking options: A) 0, B) 3tan2θ, C) 3, D) 3+tan2θ. For θ=0°, only Option C gives 3.
Expression:cos2θ3−4sin2θ + 2 tan² θ
sin2θ+cos2θ=1, tanθ=cosθsinθ, sec2θ=1+tan2θ
Substitute θ=0°. The expression becomes 13−0+0=3. Checking options: A) 0, B) 3tan2θ, C) 3, D) 3+tan2θ. For θ=0°, only Option C gives 3.
Students often struggle to expand the numerator or simplify the fraction, forgetting that cos2θ1 is sec2θ.
Expand the fraction
Divide the terms in the numerator by the denominator cos2θ.
cos2θ3−cos2θ4sin2θ+2tan2θ
Use identity for tangent
Apply the identity tanθ=cosθsinθ and secθ=cosθ1.
3sec2θ−4tan2θ+2tan2θ
Simplify terms
Combine the tan2θ terms: −4tan2θ+2tan2θ=−2tan2θ.
3sec2θ−2tan2θ
Apply secant-tangent identity
Use the identity sec2θ=1+tan2θ.
3(1+tan2θ)−2tan2θ=3+3tan2θ−2tan2θ=3+tan2θ
Option C is correct because the expression simplifies to 3+tan2θ.
This type of simplification frequently appears in SSC and RRB exams; always check simple angles like 0° or 45° to verify expressions quickly.