X = n\(\pi\) -\({\pi\over 6},\)\(n \in z\)

\(\theta ={n\pi\over3}+{\pi\over12},n \in z\)

\(\theta ={n\pi\over2}+{\pi\over6},n \in z\)

We have 
L.H.S. =\(sin^2{\pi\over6}+cos^2{\pi\over3}-tan^2{\pi\over 4}\)
\(=({1\over2})^2+({1\over2})^2-(1)^2={1\over4}+{1\over4}-1\)
\(={1+1-4\over4}={-2\over4}={-1\over2}=R.H.S\)

-225^o

\(-({1260\over 11})\)

240^o
We know that 180° = π radian
=\(({240\times {\pi\over 180}})^C=({4\pi\over 3})^C\)

- 47° 30' = -\((47{30\over 60})^c=-({95\over 2})^o\)
\(=-({95\over 2}\times{\pi\over 180})^c=-({19\pi\over 72})^c\)

\(Cos\quad A=\sqrt { 1-si{ n }^{ 2 } } A\)
\(=\sqrt { 1-\cfrac { 9 }{ 25 } =\sqrt { \cfrac { 16 }{ 25 } =\cfrac { 4 }{ 5 } } } \)
\(\Rightarrow tanA=\cfrac { 3 }{ 4 } \)
\(Also,sinB=-\sqrt { 1-co{ s }^{ 2 } } B\)
\(=-\sqrt { 1-\cfrac { 144 }{ 169 } =-\sqrt { \cfrac { 25 }{ 169 } } =-\cfrac { 5 }{ 13 } } \)
\(=tan B=\cfrac { 5 }{ 12 } \)
\(Now, tan(A-B)=\cfrac { tanA-tanB }{ 1+tanAtanB } =\cfrac { 16 }{ 63 } \)
\(and\quad cot(A+B)=1\)
\(\Rightarrow \frac { cotAcotB-1 }{ cotB+cotA } =1\)
\(\Rightarrow cotAcotB-=cotA+cotB\)
\(\Rightarrow cotAcotB-cot\quad A-cotB=1\)
\( \Rightarrow cotAcotB-cotA-cotB+1=2\)
\(\Rightarrow cotA(cotB-1)-(cotB-1)=2\)
\(\Rightarrow (cotB-1)(cotA-1)=2\)

\(\sqrt { \cfrac { 1-sin\theta }{ 1+sin\theta } } =\sqrt { \cfrac { (1-sin\theta )(1-sin\theta ) }{ (1+sin\theta )(1-sin\theta } } \)
\(=\sqrt { \cfrac { (1-sin\theta { ) }^{ 2 } }{ 1-si{ n }^{ 2 } } } =\cfrac { 1-sin\theta }{ \sqrt { co{ s }^{ 2 }\theta } }\)
\(=\cfrac { 1-sin\theta }{ |cos\theta | } \)
\(=\left\{\begin{array}{l} \frac{1-\sin \theta}{\cos \theta}, \text { if }-\frac{\pi}{2}<\theta<\frac{\pi}{2} \\ \frac{1-\sin \theta}{-\cos \theta}, \text { if } \frac{\pi}{2}<\theta<\frac{3 \pi}{2} \end{array}\right.\)

(cos 55⁰+cos65⁰+cos75⁰)+ cos 75⁰
=2cos 60⁰ cos 5⁰+cos75⁰
=cos 5⁰ + cos75⁰= 2cos40⁰cos 35⁰

\(\because (cos\theta +sin\theta )^{ 2 }=(\sqrt { 2 } cos\theta )^{ 2 }\)
\(\Rightarrow 1+2sin\theta cos\theta =2cos^{ 2 }\theta \)
\(\Rightarrow 1+2sin\theta cos\theta =2-2sin^{ 2 }\theta\)
\(\Rightarrow 2{ sin }^{ 2 }\theta =1-2sin\theta cos\theta\)
\(\Rightarrow 2{ sin }^{ 2 }\theta ={ sin }^{ 2 }\theta +{ cos }^{ 2 }\theta -2sin\theta cos\theta \)

Here,
\(\theta =3{ 0 }^{ 0 }=(30\times \cfrac { \pi }{ 180 } { ) }^{ c }=(\cfrac { \pi }{ 6 } { ) }^{ c }\quad and\quad arc=50\quad m\)
\(\because \theta =\cfrac { Arc }{ Radius } \Rightarrow \cfrac { \pi }{ 6 } =\cfrac { 50 }{ Radius } \)
\(\therefore Radius\quad =\cfrac { 50\times 6\times 7 }{ 22 } =95.45\quad m\)

Here, x+ 2x+3x=180⁰ \(\Rightarrow \)x=30⁰
\(\therefore \) \(\angle \)A=30⁰, \(\angle \)B=60⁰, \(\angle \)C=90⁰
\(\therefore \)\(\angle \)A:\(\angle \)B:\(\angle \)C=1:\(\sqrt { 3 } \):2

\(LHS\quad =\quad sin\quad (n+1)x\quad sin\quad (n+2)x\quad +\quad cos\quad (n+1)x.cos\quad (n+2)x\)
\(= cos\quad (n+1)x.cos(n+2)x+sin(n+1)x.sin(n+2)x\)
\(= cos\quad [(n+1)x-(n+2)x]\)