\(8\sqrt { 3 } { cosec }^{ 2 }{ 30 }^{ 0 }\sin { { 60 }^{ 0 } } \cos { { 60 }^{ 0 } } { cos }^{ 2 }{ 45 }^{ 0 }\sin { { 45 }^{ 0 } } \tan { { 30 }^{ 0 } } { cosec }^{ 3 }{ 45 }^{ 0 }\)
\(=8\sqrt { 3 } .\frac { 1 }{ \sin ^{ 2 }{ { 30 }^{ 0 } } } .\sin { { 60 }^{ 0 } } .\cos { { 60 }^{ 0 } } .{ cos }^{ 2 }{ 45 }^{ 0 }.\sin { { 45 }^{ 0 } } .\frac { \sin { { 30 }^{ 0 } } }{ \cos { { 30 }^{ 0 } } } .\frac { 1 }{ \sin ^{ 3 }{ { 45 }^{ 0 } } } \)
\(=8\sqrt { 3 } .\frac { 1 }{ \sin ^{ 2 }{ { 30 }^{ 0 } } } .\sin { { 60 }^{ 0 } } .\cos { { 60 }^{ 0 } } .{ cos }^{ 2 }{ 45 }^{ 0 }.\sin { { 45 }^{ 0 } } .\frac { 1 }{ \cos { { 30 }^{ 0 } } } .\frac { 1 }{ \sin ^{ 2 }{ { 45 }^{ 0 } } } \)
\(=8\sqrt { 3 } .\frac { 1 }{ 1/2 } \frac { \sqrt { 3 } }{ 2 } .\frac { 1 }{ 2 } { \left( \frac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }.\frac { 1 }{ \sqrt { 3 } /2 } .\frac { 1 }{ { (1/\sqrt { 2 } ) }^{ 2 } } \)
\(=8\sqrt { 3 } .\frac { 1 }{ 4 } .\frac { 2 }{ \sqrt { 3 } } .2=\frac { 24 }{ \sqrt { 3 } } =\frac { 24\times \sqrt { 3 } }{ \sqrt { 3 } \times \sqrt { 3 } } \)   [on multiplying numerator and denominator by \(\sqrt { 3 } \)]
\(=\frac { 24\times \sqrt { 3 } }{ 3 } =8\sqrt { 3 } \) .

since, sin (A - B) = \(\frac{1}{2}\), therefore, A - B = 30°              (1)
Also, since cos (A + B) = \(\frac{1}{2}\), therefore, A + B = 60°   (2)
Solving (1) and (2), we get : A = 45° and B = 15°.

Since we will apply the identity involving sec \(\theta\) and tan \(\theta\), let us first convert the LHS (of the identity we need to prove) in terms of sec \(\theta\) and tan \(\theta\) by dividing numerator and denominator by cos \(\theta\).
\(\begin{aligned} \text { LHS } & =\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{(\tan \theta+\sec \theta)-1}{(\tan \theta-\sec \theta)+1}=\frac{\{(\tan \theta+\sec \theta)-1\}(\tan \theta-\sec \theta)}{\{(\tan \theta-\sec \theta)+1\}(\tan \theta-\sec \theta)} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{\left(\tan ^2 \theta-\sec ^2 \theta\right)-(\tan \theta-\sec \theta)}{\{\tan \theta-\sec \theta+1\}(\tan \theta-\sec \theta)} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{-1-\tan \theta+\sec \theta}{(\tan \theta-\sec \theta+1)(\tan \theta-\sec \theta)} \end{aligned}\)
\(=\frac{-1}{\tan \theta-\sec \theta}=\frac{1}{\sec \theta-\tan \theta}\)
which is the RHS of the identity, we are required to prove.

LHS = \(\sqrt { \frac { 1-\cos { \theta } }{ 1+\cos { \theta } } } =\sqrt { \frac { 1-\frac { 1 }{ 2 } }{ 1+\frac { 1 }{ 2 } } } \left( \because \quad \cos { { 60 }^{ ° } } =\frac { 1 }{ 2 } \right) \)
\(=\sqrt { \frac { \frac { 1 }{ 2 } }{ \frac { 3 }{ 2 } } } =\frac { 1 }{ \sqrt { 3 } } \)
RHS \(=\frac { \sin { \theta } }{ 1+\cos { \theta } } =\frac { \sin { { 60 }^{ ° } } }{ 1+\cos { { 60 }^{ ° } } } \)
\(=\frac { \frac { \sqrt { 3 } }{ 2 } }{ 1+\frac { 1 }{ 2 } } =\frac { \frac { \sqrt { 3 } }{ 2 } }{ \frac { 3 }{ 2 } } \)
\(=\frac { 1 }{ \sqrt { 3 } } =LHS\)
Hence relation is verified for \(\theta ={ 60 }^{ ° }\)

\(LHS=\frac { \cos { A } }{ 1-\tan { A } } +\frac { \sin { A } }{ 1-\cot { A } } \)
\(=\frac { \cos { A } }{ 1-\left( \frac { \sin { A } }{ \cos { A } } \right) } +\frac { \sin { A } }{ 1-\left( \frac { \cos { A } }{ \sin { A } } \right) } \)
\(=\frac { \cos ^{ 2 }{ A } }{ \cos { A } .\sin { A } } +\frac { \sin ^{ 2 }{ A } }{ \sin { A } .\cos { A } } \)
\(=\frac { \cos ^{ 2 }{ A } }{ \cos { A } -\sin { A } } .\frac { \sin ^{ 2 }{ A } }{ \cos { A } -\sin { A } } \)
\(=\frac { \cos ^{ 2 }{ A.\sin ^{ 2 }{ A } } }{ \cos { A } .\sin { A } } \)
\(=\frac { \left( \cos { A } -\sin { A } \right) \left( \cos { A } +\sin { A } \right) }{ \left( \cos { A } -\sin { A } \right) } \)
\(=\cos { A } +\sin { A } \)
= RHS

\(LHS=\frac { \cos { A } }{ 1+\tan { A } } -\frac { \sin { A } }{ 1+\cot { A } } \)
\(=\frac { \cos { A } }{ 1+\frac { \sin { A } }{ \cos { A } } } -\frac { \sin { A } }{ 1+\frac { \cos { A } }{ \sin { A } } } \)
\(=\frac { \cos ^{ 2 }{ A } }{ \cos { A } +\sin { A } } -\frac { \sin ^{ 2 }{ A } }{ \sin { A+\cos { A } } } \)
\(=\frac { \cos ^{ 2 }{ A-\sin ^{ 2 }{ A } } }{ \left( \sin { A+\cos { A } } \right) } \)
\(=\frac { \left( \cos { A } +\sin { A } \right) \left( \cos { A } -\sin { A } \right) }{ \sin { A+\cos { A } } } \)
\(=\cos { A } -\sin { A } \)
= RHS

\(\frac { \sin { \theta } .2\sin ^{ 3 }{ \theta } }{ 2\cos ^{ 3 }{ \theta } .\cos { \theta } } =\frac { \sin { \theta } \left( 1-2\sin ^{ 2 }{ \theta } \right) }{ \cos { \theta } \left( 2\cos ^{ 2 }{ \theta } -1 \right) } \)
\(=\frac { \sin { \theta } \left( \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } -2\sin ^{ 2 }{ \theta } \right) }{ \cos { \theta } \left( 2\cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } -\cos ^{ 2 }{ \theta } \right) } \)
\(=\frac { \tan { \theta } \left( \cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } \right) }{ \left( \cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } \right) } \)
\(=\tan { \theta } \)

\(LHS=\left( cosec\theta -\sin { \theta } \right) \left( \sec { \theta } -\cos { \theta } \right) \left( \tan { \theta } +\cot { \theta } \right) \)
\(=\left( \frac { 1 }{ \sin { \theta } } -\sin { \theta } \right) \left( \frac { 1 }{ \cos { \theta } } -\cos { \theta } \right) \left( \frac { \sin { \theta } }{ \cos { \theta } } +\frac { \cos { \theta } }{ \sin { \theta } } \right) \)
\(=\left( \frac { 1-\sin ^{ 2 }{ \theta } }{ \sin { \theta } } \right) \left( \frac { 1-\cos ^{ 2 }{ \theta } }{ \cos { \theta } } \right) \left( \frac { \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } }{ \sin { \theta } .\cos { \theta } } \right) \)
\(=\frac { \cos ^{ 2 }{ \theta } }{ \sin { \theta } } \times \frac { \sin ^{ 2 }{ \theta } }{ \cos { \theta } } \times \left( \frac { 1 }{ \sin { \theta } .\cos { \theta } } \right) \)
\(\left[ \because \quad \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)
= 1
= RHS

sin 60° cos 30° + sin 30° cos 60°
\(=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{1}{2}\)
\(\left[\because \sin 60^{\circ}=\cos 30^{\circ}=\frac{\sqrt{3}}{2} \text { and } \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}\right]\)
\(=\frac{3}{4}+\frac{1}{4}=\frac{3+1}{4}=\frac{4}{4}=1\)

2tan²45° + cos²30° − sin²60°
\(=2(1)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}\)
\(=2(1)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}\)
\(=2+\frac{3}{4}-\frac{3}{4}=2\) 

\(\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\)
\(=\frac{\frac{1}{2}+\frac{1}{1}-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+\frac{1}{1}}=\frac{\frac{\sqrt{3}+2 \sqrt{3}-4}{2 \sqrt{3}}}{\frac{4+\sqrt{3}+2 \sqrt{3}}{2 \sqrt{3}}}\)
\(\begin{aligned} & {\left[\begin{array}{c} \because \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}, \operatorname{cosec} 60^{\circ}=\sec 30^{\circ}=\frac{2}{\sqrt{3}}, \\ \cot 45^{\circ}=\tan 45^{\circ}=1 \end{array}\right]} \\ \end{aligned}\)
\(=\frac{3 \sqrt{3}-4}{4+3 \sqrt{3}}=\frac{3 \sqrt{3}-4}{4+3 \sqrt{3}} \times \frac{4-3 \sqrt{3}}{4-3 \sqrt{3}}\)
[multiplying numerator and denominator by the conjugate of 4 + 3\(\sqrt3\), i.e. 4 - 3\(\sqrt3\)]
\(\begin{aligned} & =\frac{12 \sqrt{3}-27-16+12 \sqrt{3}}{(4)^2-(3 \sqrt{3})^2}\left[\because(a+b)(a-b)=a^2-b^2\right] \\ \end{aligned}\)
\(\begin{aligned} =\frac{24 \sqrt{3}-43}{16-27}=\frac{-(43-24 \sqrt{3})}{-11}=\frac{43-24 \sqrt{3}}{11} \end{aligned}\)

\(\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}\)
\(\frac{5\left(\frac{1}{2}\right)^{2}+4\left(\frac{2}{\sqrt{3}}\right)^{2}-(1)^{2}}{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\)
\(=\frac{5\left(\frac{1}{4}\right)+\left(\frac{16}{3}\right)-1}{\frac{1}{4}+\frac{3}{4}}\)
\(=\frac{\frac{15+64-12}{12}}{\frac{4}{4}}=\frac{67}{12}\) 

LHS = sec A (1 – sin A)(sec A + tan A) \(=\left(\frac{1}{\cos A}\right)(1-\sin A)\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)\)
\(=\frac{(1-\sin \mathrm{A})(1+\sin \mathrm{A})}{\cos ^{2} \mathrm{~A}}=\frac{1-\sin ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}\)
\(=\frac{\cos ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}=1=\mathrm{RHS}\) 

Given, cos A = \(\frac{5}{13}=\frac{\operatorname{Base}(B)}{\text { Hypotenuse }(H)}\)
By Pythagoras theorem, we get
H² = P² + B²
\(\Rightarrow\) (13)² = P² + (5)²
\(\Rightarrow\) 169 = P² + 25
\(\Rightarrow\)  169 - 25 = P²
\(\Rightarrow\)   144 = P²
\(\Rightarrow\)     P = 12
Now, we have to verify that
\(\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\cos A+\sin A\)
\(\begin{aligned} \text { LHS } & =\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A} \end{aligned}\)
\(\begin{aligned} =\frac{\frac{5}{13}}{1-\frac{12}{5}}+\frac{\frac{12}{13}}{1-\frac{5}{12}} \end{aligned}\)
\(=\frac{\frac{5}{13}}{\frac{5-12}{5}}+\frac{\frac{12}{13}}{\frac{12-5}{12}}=\frac{5}{13} \times\left(-\frac{5}{7}\right)+\frac{12}{13} \times \frac{12}{7}\)
\(=-\frac{25}{91}+\frac{144}{91}=\frac{119}{91}=\frac{17}{13}\)
RHS = cosA + sinA
\(=\frac{5}{13}+\frac{12}{13}=\frac{17}{13}\)    
LHS = RHS            Hence proved.

\(\frac { { tan }^{ 2 }60°+{ 4sin }^{ 2 }45°+{ 3sec }^{ 2 }30°+{ 5cos }^{ 2 }90° }{ cosec30°+sec60°-{ cot }^{ 2 }30° } \)
=\(\frac { { \left( 3 \right) }^{ 2 }+4\times { \left( \frac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+3\times { \left( \frac { 2 }{ \sqrt { 3 } } \right) }^{ 2 }+5\times 0 }{ 2+2-{ \left( \sqrt { 3 } \right) }^{ 2 } } \)
=\(\frac{3+2+4}{1}\)=9

\( \frac{3}{2}\) 

\(\frac{29 \sqrt{3}+1}{4 \sqrt{3}}\)

\(\frac{3(1-\sqrt{3})}{2}\)

x = 1

A = 60° , B = 30°

Convert sec x and tan x into sin and cos using, \(\sec x=\frac{1}{\cos x} \text { and } \tan x=\frac{\sin x}{\cos x} \) in both LHS and RHS separately and simplify.

\(\because\) We know that
\(\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \sin 60^{\circ}=\frac{\sqrt{3}}{2}, \sec 30^{\circ}=\frac{2}{\sqrt{3}} \text { and } \operatorname{cosec} 30^{\circ}=2\)
Now, \(\frac{\cos 45^{\circ}+\sin 60^{\circ}}{\operatorname{sec} 30^{\circ}+\operatorname{cosec} 30^{\circ}}=\frac{\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}}{\frac{2}{\sqrt{3}}+2}=\frac{\frac{2+\sqrt{6}}{2 \sqrt{2}}}{\frac{2+2 \sqrt{3}}{\sqrt{3}}}\)
\(=\frac{(2+\sqrt{6}) \sqrt{3}}{2 \sqrt{2}(2+2 \sqrt{3})}=\frac{2 \sqrt{3}+\sqrt{18}}{4 \sqrt{2}+4 \sqrt{6}}\)
\(=\frac{2 \sqrt{3}+3 \sqrt{2}}{4 \sqrt{2}+4 \sqrt{6}}=\frac{2 \sqrt{3}+3 \sqrt{2}}{4 \sqrt{2}+4 \sqrt{6}} \times \frac{4 \sqrt{2}-4 \sqrt{6}}{4 \sqrt{2}-4 \sqrt{6}}\)
\(=\frac{8 \sqrt{6}-8 \sqrt{18}+12 \times 2-12 \sqrt{12}}{32-96}\)
\(=\frac{8 \sqrt{6}-24 \sqrt{2}+24-24 \sqrt{3}}{-64}=\frac{-\sqrt{6}+3 \sqrt{2}-3+3 \sqrt{3}}{8}\)
\(=\frac{+\sqrt{2}(-\sqrt{3}+3)+\sqrt{3}(-\sqrt{3}+3)}{8}=\frac{(-\sqrt{3}+3)(\sqrt{2}+\sqrt{3})}{8}\)
\(\therefore \frac{\cos 45^{\circ}+\sin 60^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}=\frac{(-\sqrt{3}+3)(\sqrt{2}+\sqrt{3})}{8}\)

To find \(\frac{5 \tan 60^{\circ}}{\left(\sin ^2 60^{\circ}+\cos ^2 60^{\circ}\right) \tan 30^{\circ}}\)         ...(i)
We have, \(\tan 60^{\circ}=\sqrt{3}, \tan 30^{\circ}=\frac{1}{\sqrt{3}}, \sin 60^{\circ}=\frac{\sqrt{3}}{2}\)
and \(\cos 60^{\circ}=\frac{1}{2}\)
On substituting the values of tan60°, tan 30° and sin60° and cos60° in Eq. (i), we get
\(\begin{gathered} \frac{5 \times \sqrt{3}}{\left[\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2\right] \times \frac{1}{\sqrt{3}}} \\ \end{gathered}\)
\(\begin{gathered} =\frac{5 \sqrt{3}}{\left(\frac{3}{4}+\frac{1}{4}\right) \times \frac{1}{\sqrt{3}}} \end{gathered}\)
\(\Rightarrow \quad 5 \sqrt{3} \times \sqrt{3}=15\)
Therefore, \(\frac{5 \tan 60^{\circ}}{\left(\sin ^2 60^{\circ}+\cos ^2 60^{\circ}\right) \tan 30^{\circ}}=15\)