(a) 0
(b) 1/3
(c) 2/3
(d) `frac{3}{4}`![]()
(c) 2/3
(d) `frac{3}{4}`
(a) 0
Explanation:
Given,
4 tanΞ² = 3
tanΞ² = `3/4`
Now,
`"π πππ½−3 cosπ½"/"4 sin π½+3 cos"`
= `"3-3"/"3+3"`
= `"0"/"6"`
= 0
= `"0"/"6"`
= 0
If 4 tanΞ² = 3, then 4 `"π πππ½−3 cosπ½"/"4 sin π½+3 cos"` - Bzziii.com
If 4 tanΞ² = 3, then 4 `"π πππ½−3 cosπ½"/"4 sin π½+3 cos"`
(a) 0
(a) 0
Explanation:
Given,
4 tanΞ² = 3
tanΞ² = `3/4`
Now,
`"π πππ½−3 cosπ½"/"4 sin π½+3 cos"`
![]()
(a) 0
(a) 0
Explanation:
Given,
4 tanΞ² = 3
tanΞ² = `3/4`
Now,
`"π πππ½−3 cosπ½"/"4 sin π½+3 cos"`
![](https://www.pinterest.com/pin/create/button/?url=https://bzzii.blogspot.com/2021/09/if-4-tan-3-then-4-3-cos4-sin-3-cos.html&media=https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjfEdj-OaK9iJLkfYnb4edgAWZ3KETfNKF0gAlaQ0vQWfyvyof6xgf1yoNzTsM7ojo4LJhX2xd3qex0st6ToKxDT_SyNF38CnDDlJ2dEOXLgGL06rWrz6SCXBAEa3-CByEVoJLtI8pK6Lg/w319-h226/a.png&description=If 4 tanΞ² = 3, then 4 `"π πππ½β3 cosπ½"/"4 sin π½+3 cos"` - Bzziii.com){kind=link}
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