VSA2 marksYear: 2023EasyCBSE Class 10
trigonometrystandard-anglestrigonometric-ratios
Question
If 4cot245°−sec260°+sin260°+p=43, then find the value of p.
Solution
1
4cot245°−sec260°+sin260°+p=43
⇒4(1)2−(2)2+(23)2+p=43 +1 mark2
⇒4−4+43+p=43⇒p=0
+1 mark