Trigonometry — BSSC Quantitative Aptitude

1. $\sec A = \frac{13}{5} \implies \cos A = \frac{5}{13}$. 2. Using Pythagorean triplet $(5, 12, 13)$, Perpendicular $P = 12$. 3. Divide numerator and denominator by $\sin A$: $$\frac{\tan A + \sin A}{\tan A - \sin A} = \frac{\sec A + 1}{\sec A - 1} = \frac{\frac{13}{5} + 1}{\frac{13}{5} - 1} = \frac{18/5}{8/5} = \frac{9}{4}$$

1. Perpendicular $P = 2xy$ and Hypotenuse $H = x^2 + y^2$. 2. Base $B = \sqrt{H^2 - P^2} = \sqrt{(x^2+y^2)^2 - (2xy)^2} = \sqrt{(x^2-y^2)^2} = x^2 - y^2$. 3. $\tan \theta = \frac{P}{B} = \frac{2xy}{x^2 - y^2}$.

1. Squaring both sides: $(\sin A + \cos A)^2 = 2 \implies 1 + 2\sin A\cos A = 2 \implies \sin A\cos A = \frac{1}{2}$. 2. $\sin^4 A + \cos^4 A = (\sin^2 A + \cos^2 A)^2 - 2\sin^2 A\cos^2 A = 1 - 2\left(\frac{1}{2}\right)^2 = 1 - \frac{1}{2} = \frac{1}{2}$.

1. Since $\sec^2 \theta - \tan^2 \theta = 1 \implies \sec \theta - \tan \theta = \frac{1}{5}$. 2. Adding: $2\sec \theta = 5 + \frac{1}{5} = \frac{26}{5} \implies \sec \theta = \frac{13}{5} \implies \cos \theta = \frac{5}{13}$. 3. Subtracting: $2\tan \theta = \frac{24}{5} \implies \tan \theta = \frac{12}{5}$. 4. $\sin \theta = \frac{\tan \theta}{\sec \theta} = \frac{12/5}{13/5} = \frac{12}{13}$.

1. Identity: $(a\sin X + b\cos X)^2 + (a\cos X - b\sin X)^2 = a^2 + b^2$. 2. Here $a = 3$, $b = 5$, and $3\sin \theta + 5\cos \theta = 5$. 3. $5^2 + (3\cos \theta - 5\sin \theta)^2 = 3^2 + 5^2 \implies (3\cos \theta - 5\sin \theta)^2 = 9$. 4. $3\cos \theta - 5\sin \theta = 3$ (positive value).

1. Complementary relation: $\sin A = \cos B \implies A + B = 90^\circ$. 2. $(2x + 15^\circ) + (x - 5^\circ) = 90^\circ$. 3. $3x + 10^\circ = 90^\circ \implies 3x = 80^\circ \implies x = 26.67^\circ$.

1. $\tan 80^\circ = \cot 10^\circ$ and $\tan 70^\circ = \cot 20^\circ$. 2. $(\tan 10^\circ \cot 10^\circ) \cdot (\tan 20^\circ \cot 20^\circ) \cdot \tan 30^\circ$. 3. $1 \cdot 1 \cdot \tan 30^\circ = \frac{1}{\sqrt{3}}$.

1. Denominator = $\sin^2 30^\circ + \cos^2 30^\circ = 1$. 2. Numerator = $5(1/2)^2 + 4(2/\sqrt{3})^2 - 1^2 = \frac{5}{4} + \frac{16}{3} - 1 = \frac{15 + 64 - 12}{12} = \frac{67}{12}$.

1. In right-angled triangle $\Delta ABC$ at $C$, $A + B = 90^\circ \implies B = 90^\circ - A$. 2. $\cos B = \cos(90^\circ - A) = \sin A$. 3. Given $\sin A = \frac{3}{5}$, therefore $\cos B = \frac{3}{5}$.

1. $\tan\theta + \cot\theta = 2 \implies \theta = 45^\circ$ (since $\tan 45^\circ = 1$ and $\cot 45^\circ = 1$). 2. Substituting $\tan\theta = 1$ and $\cot\theta = 1$: $$1^3 + 1^3 + 2(1^5)(1^7) = 1 + 1 + 2 = 4$$