Trigonometric Functions — MATHS STD 11 Science — Question

1. $1$

Solution:

We have,
$\Rightarrow2\cos^2​​\text{x}-1+2\cos\text{x}=1$

$\Rightarrow\cos^2\text{x}+\cos\text{x}-1=0$

$\Rightarrow\cos\text{x}=\frac{-1\pm\sqrt{1^2+4}}{2}$

$\Rightarrow\cos\text{x}=\frac{-1\pm\sqrt{5}}{2}$

$\Rightarrow\cos\text{x}=\frac{-1+\sqrt{5}}{2}$

Now,

$(2-\cos^2​​\text{x})\sin^2\text{x}=\bigg[2-\Big(\frac{-1+\sqrt{5}}{2}\Big)^2\bigg](1-\cos^2\text{x})$

$=\bigg[2-\frac{1}{4}\Big(1-2\sqrt{5}+5\Big)\bigg]\Big(1-\frac{1}{4}\Big(1-2\sqrt{5}+5\Big)\Big)$

$=\frac{1}{4}\Big(1+\sqrt{5}\Big)\Big(\sqrt{5}-1\Big)=\frac{4}{4}=1$