The value of `cos^2(pi/6+theta)-sin^2\ \ (pi/6-theta)` is `1/2cos2theta` b. `0` c. `-1/2cos2theta` d. `-cot35^0`

Explanation

`cos ^2(π/6 + θ) – sin^2(π/6 – θ)`



Using the formula `cos ^2A – sin^2B = cos(A + B) cos(A – B)`,



= `cos(π/6 + θ + π/6 – θ) cos(π/6 + θ – π/6 + θ)`



= `cos(2π/6) cos^ 2θ`



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