If 4 cot^2 45° — sec2^ 60° + sin^2 60° + p = 3/4 then find the value

Q) If 4 cot²45° – sec² 60° + sin² 60° + p = $\frac{3}{4}$ then find the value of p.

Ans: 4 cot²45° – sec² 60° + sin² 60° + p = $\frac{3}{4}$

4 (1)² – (2)² + $(\frac{\sqrt 3}{2})^2 + p = \frac{3}{4}$

4 – 4 + $\frac{3}{4}$ + p = $\frac{3}{4}$

$\therefore$ p = 0

Therefore, the value of p is 0

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