\(\cfrac { cos{ 63 }^{ 0 }20' }{ sin{ 26 }^{ 0 }40' } =\cfrac { cos63^{ 0 }20' }{ cos{ 63 }^{ 0 }20' } =1\)

\( \frac{\tan 25^{\circ}}{\cot 65^{\circ}}+\frac{\sin 40^{\circ}}{\cos 50^{\circ}} \)
\(= \frac{\tan \left(90^{\circ}-65^{\circ}\right)}{\cot 65^{\circ}}+\frac{\sin \left(90^{\circ}-50^{\circ}\right)}{\cos 50^{\circ}} \)
\(= \frac{\cot 65^{\circ}}{\cot 65^{\circ}}+\frac{\cos 50^{\circ}}{\cos 50^{\circ}}=1+1=2 \)

cot (90° - 75) cot (90° - 60°) cot 45° cot 60° cot 75°
= tan 75° tan 6.0° (1) cot 60° cot 75° = 1

3 tan \(\theta\) + 4 (sec \(\theta\) \(\times\) sin\(\theta\)) = 24
\(3tan\theta +4\left( \cfrac { 1 }{ cos\theta } \times sin\theta \right) =24\)
7 tan \(\theta\) = 24
\(tan\theta =\cfrac { 24 }{ 7 } \)
hypetenuse = \(\sqrt { { 24 }^{ 2 }+49 } =\sqrt { 576+49 } =\sqrt { 625 } =25\)
\(sin\theta =\cfrac { 24 }{ 25 } ;cos\theta =\cfrac { 7 }{ 25 } ;cosec\theta =\cfrac { 25 }{ 24 } ;sec\theta =\cfrac { 25 }{ 27 } ;cot\theta =\cfrac { 7 }{ 24 } \)

,Adjacent side = \(\sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+2bc-{ a }^{ 2 } } \)
= \(\sqrt { \left( b+c \right) ^{ 2 } } =b+c\)
\(\therefore\) \((b+c)sin\theta =\alpha cos\theta \)
\(\left( b+c \right) \times \cfrac { a }{ \sqrt { \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+2bc \right) } } =a\times \cfrac { b+c }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+2bc } } \)

\(cot\theta =1\)
\(cot\theta =\cfrac { 1 }{ 3 } \)
\(\cfrac { adjacent }{ opposite } =\cfrac { 1 }{ 3 } \)
\(\sqrt { { 3 }^{ 2 }+1 } =\sqrt { 10 } \)
\(\cfrac { 3cos\theta -4sin\theta }{ 5sin\theta +4cos\theta } =\cfrac { 3\times \cfrac { 1 }{ \sqrt { 10 } } -4\times \cfrac { 3 }{ \sqrt { 10 } } }{ 5\times \cfrac { 3 }{ \sqrt { 10 } } +4\times \cfrac { 1 }{ \sqrt { 10 } } } =\cfrac { 3-12 }{ 15+4 } =\cfrac { -9 }{ 19 } \)

\(cos\theta =\cfrac { 4 }{ 5 } =0.8\)
cos 36° 48' = 0.8
\(\therefore\) \(\theta\) = 36⁰48'

Hypotenuse = 10 cm
One of the acute angle = 24° 24'
sin 24° 24' = 0.4131'
\(\cfrac { x }{ 10 } =0.4131\)
x = 0.4131 \(\times\) 10
x = 4.131
cos 24° 24' = 0.9107
\(\cfrac { y }{ 10 } =0.9107\)
y = 9.107
\(\therefore\) Area of the triangle = \(\cfrac { 1 }{ 2 } bh\)
= \(\cfrac { 1 }{ 2 } \times y\times x\)
\(=\cfrac { 1 }{ 2 } \times 9.107\times 4.131=18.81sq.cm\)

(i) From the natural sines table
sin 85° 57' = 0.9975
\(\therefore\) \(\theta \) = 85° 57'
(ii) cos\(\theta \) = 0.6763
cos 47° 33' = 0.6762
\(\therefore\) \(\theta \) = 47° 33'
(iii) tan \(\theta \) = 0.0720
tan 4° 7' = 0.0720
\(\therefore\)\(\theta \) = 4°7'
(iv) cos \(\theta \) = 0.0410
cos 87° 45° = 0.0410
:\(\therefore\)\(\theta \) = 87°39'
(v) tan \(\theta \) = 7.5958
tan 82° 30' = 7.5958
\(\therefore\) \(\theta \) = 82° 30'

8 sin 2(15°). cos 4 (15°). sin 6(15°)
= 8 sin 30° cos 60° sin 90°
= \(8\times \cfrac { 1 }{ 2 } \times \cfrac { 1 }{ 2 } \times 1=2\)

L.H.S = cos 3A= cos 3 (30°)
= cos 90°
=0 ---- (1)
R.H.S = 4 cos³ A- 3 cosA
= 4 cos³ 30° - 3 cos 30°
= \(4\left( \cfrac { \sqrt { 3 } }{ 2 } \right) ^{ 3 }-3\times \cfrac { \sqrt { 3 } }{ 2 } \)

(1) = (2). Hence it is verified

\(sin\ \theta =\cfrac { a }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)
\(cos\ \theta =\cfrac { b }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)
\(b\ sin\ \theta =b\times \cfrac { a }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } -\cfrac { ab }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)    ... (1)
\(a\ cos\ \theta =a\times \cfrac { b }{ \sqrt { a^{ 2 }+{ b }^{ 2 } } } =\cfrac { ab }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)    ...(2)
(1) = (2)\(\Rightarrow\) LHS = RHS
Hence proved

By By the Pythagoras theorem
BC² = AC² + AB²
= \(\left( \sqrt { { a }^{ 2 }+{ b }^{ 2 } } \right) ^{ 2 }-{ a }^{ 2 }\)
= a²+b²-a²
= b²
BC = b

By the pythagoras theorem,
AB² = OA² + OB²
(1 + x²)² = (2x)² + OB²
OB² = (1 +x²)² - (2x)² = 1+ x⁴ + 2x² - 4x² = 1+x⁴ - 2x²
OB² = (1-x²)² B
OB = (1-x²)
\(\therefore sin\ A=\cfrac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \)
\(tan\ A=\cfrac { 1-{ x }^{ 2 } }{ 2x } \)

(i) \(sinB=\cfrac { 12 }{ 13 } \)
(ii) \(secB=\cfrac { 1 }{ cosB } \)
=  \(\cfrac { 1 }{ 5/3 } =\cfrac { 13 }{ 5 } \)
(iii) \(cotB=\cfrac { 1 }{ tanB } =\cfrac { 1 }{ 12/5 } =\cfrac { 5 }{ 2 } \)
(iv) 

(v) \(\tan C=\cfrac { 12 }{ 16 } =\cfrac { 3 }{ 4 } \)
(vi)  \(cosecC=\cfrac { 1 }{ sinC } =\cfrac { 1 }{ 12/20 } =\cfrac { 20 }{ 12 } =\cfrac { 5 }{ 3 } \)

By the pythagoras theorem,
\(AD=\sqrt { { 13 }^{ 2 }-{ 5 }^{ 2 } } \)
= \(\sqrt { 169-25 } \)
= \(\sqrt { 144 } =12\)
AC =\(\sqrt { { 12 }^{ 2 }+{ 16 }^{ 2 } } \)
= \(\sqrt { { 144 }+{ 256 } } \)
= \(\sqrt { 400 } =20\) 

sin B = \(\frac { 9 }{ 41 } \);
cos B = \(\frac { 40 }{ 41 } \);
tan B =\(\frac { 9 }{ 40 } \) ;
cosec B = \(\frac { 1 }{ sinB } \) = \(\frac { 41 }{ 9 } \);
sec B =\(\frac { 1 }{ cotB } \) = \(\frac { 41 }{ 40 } \);
cot B =\(\frac { 1 }{ tanB } \) = \(\frac { 40 }{ 9 } \)