If \(x \cos \alpha+y \sin \alpha=4\) is tangent to \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\), then the value of \(\alpha\) is

Given, \(x \cos \alpha+y \sin \alpha=4\) is a tangent to
\(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\) \(\Rightarrow \quad y=-x \cot \alpha+4 \operatorname{cosec} \alpha\) Here, \(m=-\cot \alpha, c=4 \operatorname{cosec} \alpha\) and \(\quad a^{2}=25, b^{2}=9\)
Now, the condition of tangent to ellipse is
\(c^{2}=a^{2} m^{2}+b^{2}\)
\(16 \operatorname{cosec}^{2} \alpha=25 \cot ^{2} \alpha+9\)
\(\Rightarrow \quad \frac{16}{\sin ^{2} \alpha}=\frac{25 \cos ^{2} \alpha}{\sin ^{2} \alpha}+9\)
\(\Rightarrow \quad 16=25 \cos ^{2} \alpha+9 \sin ^{2} \alpha\)
\(\Rightarrow \quad 9 \sin ^{2} \alpha+25-25 \sin ^{2} \alpha=16\)
\(\Rightarrow \quad-16 \sin ^{2} \alpha=-9\)
\(\Rightarrow \quad \sin ^{2} \alpha=\frac{9}{16}, \sin \alpha=3 / 4\)
\(\Rightarrow \quad \cos ^{2} \alpha=\frac{7}{16}\)
\(\Rightarrow \quad \tan ^{2} \alpha=9 / 7\)
\(\Rightarrow \quad \alpha=\tan ^{-1}(3 / \sqrt{7})\)