$$\eqalign{ & {\text{ }}\sqrt 3 \tan \theta = 3\sin \theta \cr & {\bf{Shortcut method:}} \cr & \Rightarrow {\text{ }}\sqrt 3 \frac{{\sin \theta }}{{\cos \theta }} = 3\sin \theta \cr & \Rightarrow \frac{{\sqrt 3 }}{{\cos \theta }} = 3 \cr & \Rightarrow \cos \theta = \frac{{\sqrt 3 }}{3} \cr & {\text{then perpendicular}} = \sqrt 6 \cr} $$

$$\eqalign{ & \Rightarrow \left( {{{\sin }^2}\theta - {\text{co}}{{\text{s}}^2}\theta } \right) \cr & \Rightarrow {\left( {\frac{P}{H}} \right)^2} - {\left( {\frac{B}{H}} \right)^2} \cr & \Rightarrow {\left( {\frac{{\sqrt 6 }}{3}} \right)^2} - {\left( {\frac{{\sqrt 3 }}{3}} \right)^2} \cr & \Rightarrow \frac{6}{9} - \frac{3}{9} \cr & \Rightarrow \frac{1}{3} \cr} $$

$$\eqalign{ & XZ - YZ = 2 \cr & h - P = 2 \,........{\text{(i)}} \cr & {h^2} = {(2\sqrt 6 )^2} + {P^2} \cr & {h^2} - {P^2} = {\left( {2\sqrt 6 } \right)^2} \cr & (h - P)(h + P) = 4 \times 6 \cr & (2)(h + P) = 24 \cr & h + P = 12 \,..........{\text{(ii)}} \cr & {\text{Adding eq }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & 2h = 14 \cr & h = 7 \,{\text{and }}P = 5 \cr & \therefore secX + tanX \cr & = \frac{h}{{XY}} + \frac{P}{{XY}} \cr & = \frac{7}{{2\sqrt 6 }} + \frac{5}{{2\sqrt 6 }} \cr & = \frac{{12}}{{2\sqrt 6 }} \cr & = \frac{6}{{\sqrt 6 }} \cr & = \sqrt 6 \cr} $$

$$\eqalign{ & = \frac{{2\sin \theta }}{{\cos \theta \left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right)}} \cr & = \frac{{2{\text{tan}}\theta }}{{1 + {\text{ta}}{{\text{n}}^2}\theta }} \cr & = {\sin ^2}\theta \cr} $$

$$\eqalign{ & \tan \theta = \frac{3}{4} \cr & {\text{cos2}}\theta = \frac{{1 - {\text{ta}}{{\text{n}}^2}\theta }}{{1 + {\text{ta}}{{\text{n}}^2}\theta }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1 - \frac{9}{{16}}}}{{1 + \frac{9}{{16}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\frac{7}{{16}}}}{{\frac{{25}}{{16}}}} \cr & {\text{cos2}}\theta = \frac{7}{{25}} \cr} $$

$$\eqalign{ & = \frac{{{\text{sin 6}}{5^ \circ }}}{{\cos {{25}^ \circ }}} \cr & = \frac{{{\text{sin }}\left( {{{90}^ \circ } - {\text{6}}{5^ \circ }} \right)}}{{\cos {{25}^ \circ }}} \cr & \left[ {\sin \left( {{{90}^ \circ } - \theta } \right) = \cos \theta } \right] \cr & = \frac{{\cos {{25}^ \circ }}}{{\cos {{25}^ \circ }}} \cr & = 1 \cr} $$

$$\eqalign{ & \alpha + \beta = {90^ \circ } \cr & {\text{and }}\alpha :\beta = 2:1 \cr & 2x + x = {90^ \circ } \cr & x = {30^ \circ } \cr & {\text{ }}\alpha = {60^ \circ } \cr & \beta = {30^ \circ } \cr & \frac{{{\text{cos }}\alpha }}{{\cos \beta }} \cr & = \frac{{\cos {{60}^ \circ }}}{{\cos {{30}^ \circ }}} \cr & = \frac{{\frac{1}{2}}}{{\frac{{\sqrt 3 }}{2}}} \cr & = \frac{1}{2} \times \frac{2}{{\sqrt 3 }} \cr & = \frac{1}{{\sqrt 3 }} \cr & = 1:\sqrt 3 \cr} $$

$$\sin \theta = \frac{a}{b} = \frac{{\text{P}}}{{\text{H}}}$$

$$\eqalign{ & {\text{BC}} = \sqrt {{b^2} - {a^2}} \cr & \left[ {{\text{using pythagorad theorem}}} \right] \cr & \therefore sec\theta - \cos \theta \cr & = \frac{{\text{H}}}{{\text{B}}} - \frac{{\text{B}}}{{\text{H}}} \cr & = \frac{{{\text{AC}}}}{{{\text{BC}}}} - \frac{{{\text{BC}}}}{{{\text{AC}}}} \cr & = \frac{b}{{\sqrt {{b^2} - {a^2}} }} - \frac{{\sqrt {{b^2} - {a^2}} }}{b} \cr & = \frac{{{b^2} - {{\left( {\sqrt {{b^2} - {a^2}} } \right)}^2}}}{{b\sqrt {{b^2} - {a^2}} }} \cr & = \frac{{{b^2} - {b^2} + {a^2}}}{{b\sqrt {{b^2} - {a^2}} }} \cr & = \frac{{{a^2}}}{{b\sqrt {{b^2} - {a^2}} }} \cr} $$

$$\eqalign{ & sec\alpha + \tan \alpha = 2\,.....(i) \cr & sec\alpha - \tan \alpha = \frac{1}{2}\,.....(ii) \cr & {\text{Adding equation (i) and (ii)}} \cr & {\text{2sec}}\alpha = 2 + \frac{1}{2} \cr & \sec \alpha = \frac{{5 \to {\text{H}}}}{{4 \to {\text{B}}}} \cr} $$

$$\sin \alpha = \frac{3}{5} = 0.6$$

$$\cos \theta = \frac{{15 \to {\text{Base}}}}{{17 \to {\text{Hypo}}}}$$

$$\eqalign{ & {\text{Perpendicular = 8}} \cr & \Rightarrow {\text{cot}}\left( {{{90}^ \circ } - \theta } \right) \cr & \Rightarrow {\text{tan}}\theta = \frac{8}{{15}}\left[ {\therefore \tan \theta = \frac{{\text{P}}}{{\text{B}}}} \right] \cr} $$

$$\eqalign{ & {\text{AC}} = \sqrt {{2^2} + {1^2}} = \sqrt 5 \cr & {\text{sin A}} + \operatorname{cotC} \cr & \frac{{{\text{BC}}}}{{{\text{AC}}}} + \frac{{{\text{BC}}}}{{{\text{AB}}}} \cr & \frac{1}{{\sqrt 5 }} + \frac{1}{2} \cr & \Rightarrow \frac{{2 + \sqrt 5 }}{{2\sqrt 5 }} \cr} $$

$$\eqalign{ & \frac{{{\pi ^c}}}{4}{\text{ = }}\frac{{{{180}^ \circ }}}{4}{\text{ = }}{45^ \circ } \cr & \angle {\text{BAC}} = {180^ \circ } - {75^ \circ } - {45^ \circ } = {60^ \circ } \cr & {180^ \circ } \to \pi \cr & {1^ \circ } \to \frac{\pi }{{{{180}^ \circ }}} \cr & {60^ \circ } \to \frac{\pi }{{{{180}^ \circ }}} \times {60^ \circ } = \frac{\pi }{3}{\text{ radian}} \cr} $$

6sin⁴θ + 3cos⁴θ = 2
⇒ 6sin⁴θ + 3(1 - sin²θ )² = 2
⇒ 6sin⁴θ + 3 + 3sin⁴θ - 6sin²θ = 2
⇒ 9sin⁴θ - 6sin²θ + 1 = 0
⇒ (3sin²θ -1)² = 0
⇒ 3sin²θ = 1
⇒ sinθ = $$\frac{1}{{\sqrt 3 }} $$


Now, $${\left[ {7{{\operatorname{cosec} }^6}\theta + 8{{\sec }^6}\theta } \right]^{\frac{1}{3}}}$$
$$\eqalign{ & = {\left[ {7 \times {{(\sqrt 3 )}^6} + 8{{\left( {\frac{{\sqrt 3 }}{{\sqrt 2 }}} \right)}^6}} \right]^{\frac{1}{3}}} \cr & = {\left[ {7 \times 27 + 8 \times \frac{{27}}{8}} \right]^{\frac{1}{3}}} \cr & = {\left( {8 \times 27} \right)^{\frac{1}{3}}} \cr & = 6 \cr} $$

$$\eqalign{ & {\text{In }}\vartriangle {\text{ABC sin2}}{{\text{1}}^ \circ }{\text{ = }}\frac{x}{y} \cr & {\text{AB}} = x \cr & {\text{AC}} = y \cr & {\text{BC}} = \sqrt {{y^2} - {x^2}} \cr & \Rightarrow {\text{sec2}}{{\text{1}}^ \circ } - \sin {69^ \circ } \cr & \Rightarrow \frac{{{\text{AC}}}}{{{\text{BC}}}} - \frac{{{\text{BC}}}}{{{\text{AC}}}} \cr & \Rightarrow \frac{{{{\left( {{\text{AC}}} \right)}^2} - {{\left( {{\text{BC}}} \right)}^2}}}{{\left( {{\text{BC}}} \right)\left( {{\text{AC}}} \right)}} = \frac{{{y^2} - {{\left( {\sqrt {\left( {{y^2} - {x^2}} \right)} } \right)}^2}}}{{y\sqrt {{y^2} - {x^2}} }} \cr & \Rightarrow \frac{{{y^2} - {y^2} + {x^2}}}{{y\sqrt {{y^2} + {x^2}} }} = \frac{{{x^2}}}{{y\sqrt {{y^2} - {x^2}} }} \cr} $$

Given, cos27° = x

tan63° = $$\frac{x}{{\sqrt {1 - {x^2}} }}$$

sin31° = $$\frac{x}{y}{\text{,}}$$
∴ sec31° - sin59°
$$\eqalign{ & = \frac{y}{{\sqrt {{y^2} - {x^2}} }} - \frac{{\sqrt {{y^2} - {x^2}} }}{y} \cr & = \frac{{{y^2} - ({y^2} - {x^2})}}{{y\sqrt {{y^2} - {x^2}} }} \cr & = \frac{{{x^2}}}{{y\sqrt {{y^2} - {x^2}} }} \cr} $$