(i) sin 38⁰36' + tan 12⁰12'
sin 38⁰36' = 0.6239
tan12⁰12' = 0.2162
sin 38⁰36' + tan 12⁰12' = 0.8401
(ii) tan 60⁰25^' - cos 49⁰20^'
tan 60⁰25' = 1.7603 + 0.0012 = 1.7615
cos 49⁰20^' = 0.6521 - 0.0004 = 0.6517
tan 60⁰25' - cos 49⁰20' = 1.1098

write 70⁰13' = 70⁰12' + 1'
From the table we have, tan70⁰12' = 2.7776 

write 19⁰59' = 19⁰54' + 5'
From the table we have, cos19⁰54' = 0.9403 

(i) We know that cosec A =  sec(90⁰ A)
sec(90⁰ - A) sec(34⁰)
90⁰ - A =34⁰
We get A = 90° − 34°
A = 56⁰
(ii) We know that tan B = cot(90⁰ - B)
cot(90⁰ - B) = cot 47⁰
90⁰ - B = 47⁰
We get B = 90° − 47°
B = 43⁰

(i) \(\frac { sin\ 49° }{ cos\ 41° } \) 
sin 49⁰ = sin(90⁰ - 41⁰) = cos 41⁰,  since 49⁰ + 41⁰ = 90⁰ (complementary),
Hence on substituting sin 49^o = cos41^o we get, \( \frac { cos\ 41° }{ cos\ 41° } \)= 1
(ii) \(\frac { sec\ 63° }{ cosec\ 27° } \)
sec63^o = sec (90^o- 27^o) = cosec27^o, here, 63^o and 27^o are complementary angles
we have \(\frac { sec\ 63° }{ cosec\ 27° } =\frac { cosec\ 27° }{ cosec\ 27° } =1\)

Let BC = 13 and AB = 5
sec θ = \(\frac { hypotenuse }{ adjacentside } =\frac { BC }{ AB } =\frac { 13 }{ 5 } \)
By the Pythagoras theorem,
\(AC=\sqrt { { BC }^{ 2 }-{ AB }^{ 2 } } \)
= \(\sqrt { { 13 }^{ 2 }-{ 5 }^{ 2 } } \)
= \(\sqrt { 169-25 } \) = \(\sqrt { 144 } \) = 12
Therefore, \(sin\theta =\frac { AC }{ BC } =\frac { 12 }{ 13 } \) ; \(cos\theta =\frac { AB }{ BC } =\frac { 5 }{ 13 } \)
\(LHS=\frac { 2sin\theta -3cos\theta }{ 4sin\theta -9cos\theta } =\frac { 2\times \frac { 12 }{ 13 } =3\times \frac { 5 }{ 13 } }{ 4\times \frac { 12 }{ 13 } -9\times \frac { 5 }{ 13 } } =\frac { \frac { 24-15 }{ 13 } }{ \frac { 48-45 }{ 12 } } =\frac { 9 }{ 3 } =3\) = RHS

tan A = \(\frac { opposite\ side }{ adjacent\ side } =\frac { 2 }{ 3 } \)
By Pythagoras theorem,
\(AC=\sqrt { { AB }^{ 2 }+{ BC }^{ 2 } } \)
= \(\sqrt { { 3 }^{ 2 }+{ 2 }^{ 2 } } =\sqrt { 9+4 } =\sqrt { 13 } \)
AC = \(\sqrt { 13 } \)
\(sin\ A=\frac { opposite\ side }{ hypotenuse } =\frac { 2 }{ \sqrt { 13 } } \)
\(cosec\ A=\frac { hypotenuse }{ opposite\ side } =\frac { \sqrt { 13 } }{ 2 } \)
\(cos\ A=\frac { adjacent\ side }{ hypotenuse } =\frac { 3 }{ \sqrt { 13 } } \)
\(sec\ A=\frac { hypotenuse }{ adjacent\ side } =\frac { \sqrt { 13 } }{ 3 } \)
\(cot\ A=\frac { adjacent\ side }{ opposite\ side }= \frac { 3 }{ 2 } \)

\(\alpha ={ sin }^{ -1 }\left( \cfrac { 1 }{ \sqrt { 2 } } \right) ={ 45 }^{ 0 }\)
\(\beta \) = 60⁰
\(\alpha +\beta ={ 105 }^{ 0 }\)

sin 3 (10°) sin 6 (10°) sin 9 (10°)
= \(\cfrac { 1 }{ 2 } \times \cfrac { \sqrt { 3 } }{ 2 } \times 1=\cfrac { \sqrt { 3 } }{4 } \)

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

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

Hypotenuse = \(\sqrt { { 12 }^{ 2 }+{ 5 }^{ 2 } } \)
= \(\sqrt { 144+25\quad } =\sqrt { 69 } \)
= 13
\(sin\theta =\cfrac { 5 }{ 13 } ;cos\theta =\cfrac { 12 }{ 13 } ;tan\theta =\cfrac { 5 }{ 12 } ; \)
\(cosec\theta =\cfrac { 13 }{ 5 } ;sec\theta =\cfrac { 13 }{ 12 } ;cot\theta =\cfrac { 12 }{ 3 } \)

If \(2cos\theta =\sqrt { 3 } \)
\(cos\theta =\cfrac { \sqrt { 3 } }{ 2 } \)
\(x=\sqrt { { 2 }^{ 2 }-\sqrt { { 3 }^{ 2 } } } =\sqrt { 4-3 } =\sqrt { 1 } =1\)
\(\therefore sin\theta =\cfrac { 1 }{ 2 } \)
\(cos\theta =\cfrac { \sqrt { 3 } }{ 2 } \)
\(tan\theta =\cfrac { 1 }{ \sqrt { 3 } } \)
\(coseec\theta =2\)
\(sec\theta =\cfrac { 2 }{ \sqrt { 3 } } \)
\(cot\theta =\sqrt { 3 } \)

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 =\(\frac { 1 }{ tanB } \) = \(\frac { 40 }{ 9 } \)

(i) sin \(\theta\) = 0.9858  = 0.9857 + 0.0001
From the sine table 0.9857 = 80^o 18'
Mean difference 1 = 2′
0.9858 = sin 80⁰20'
sin\(\theta\) = 0.9858 = sin 80⁰20'
\(\theta\) = 80⁰20'
(ii) cos \(\theta\) = 0.7656 =  0.7660 - 0.0004
From the natural cosine table
0.7660 = 40°0′
Mean difference 4 = 2′       
0.7656 = 40°0′         
cos \(\theta\) = 0.7656 = cos 40°2'
\(\theta\) = 40°2'                                                                                          

(i) (cos0⁰ + sin 45⁰ + sin 30⁰) (sin 90⁰ - cos 45⁰ + cos 60⁰)
= \(\left[ 1+\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } \right] \left[ 1-\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } \right] \)
= \(\left[ \frac { 2\sqrt { 2 } +2+\sqrt { 2 } }{ 2\sqrt { 2 } } \right] \left[ \frac { 2\sqrt { 2 } -2+\sqrt { 2 } }{ 2\sqrt { 2 } } \right] =\left[ \frac { 3\sqrt { 2 } +2 }{ 2\sqrt { 2 } } \right] \left[ \frac { 3\sqrt { 2 } -2 }{ 2\sqrt { 2 } } \right] \)
= \(\frac { 18-4 }{ 4\left( \sqrt { 2 } \right) ^{ 2 } } =\frac { 14 }{ 4\times 2 } =\frac { 7 }{ 4 } \) 
(ii) tan²60⁰ - 2tan²45⁰ - cot²30⁰ +2sin²30⁰ + \(\frac { 3 }{ 4 } \) cosec²45⁰
= \(\left( \sqrt { 3 } \right) ^{ 2 }-2(1)^{ 2 }-\left( \sqrt { 3 } \right) ^{ 2 }+2\left( \frac { 1 }{ 2 } \right) ^{ 2 }+\frac { 3 }{ 4 } \left( \sqrt { 2 } \right) ^{ 2 }\)
= \(3-2-3+\frac { 1 }{ 2 } +\frac { 3 }{ 2 } \)
= -2 + \(\frac { 4 }{ 2 } \) = -2 + 2 = 0