Question

7sin²+3cos²=4 then the value of cot​

Answer 1

We know sin²x + cos²x = 1

cotx = $\frac{cosx}{sinx}$

Coming to the question,

7sin²x+3cos²x=4

4sin²x+(3sin²x+3cos²x)=4

4sin²x+3(sin²x+cos²x)=4

4sin²x+3=4

sin²x = $\frac{1}{4}$

sinx =± $\frac{1}{2}$

Cosx = $\sqrt[]{1-sin^{2}x }$

cosx = $\sqrt{1-\frac{1}{4} }$

cosx = $\frac{\sqrt{3} }{2}$

Hence cotx = ±$\sqrt{3}$

HOPE IT HELPS !!