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```TRIGONOMETRIC RATIOS & IDENTITIES
Page # 153
Ex.12 Find satisfying the equation, tan 15Â° Â· tan 25Â° Â· tan 35Â° = tan , where
Sol. LHS = tan 15Â° Â· tan (30Â° â 5Â°) Â· tan (30Â° + 5Â°)
let t = tan 30Â° and m = tan 5Â°
tan = tan 15Â° Â·
(0, 15Â°).
t m t m
t 2 m2
3m m3 1 3m2
Â·
Â·
= tan 3(5 ) Â·
=
1 tm 1 tm
1 t 2 m2
1 3m 2 3 m 2
m (3 m 2 ) (1 3m 2 )
Â·
=
= m = tan 5Â°. Hence = 5Â°
(1 3m 2 ) 3 m 2
tan 3 x
3 tan x tan3 x
1 3 tan2 x
tan 30Âº
t 2 1/ 3
t
;
Ex.13 If tan A & tan B are the roots of the quadratic equation, a x2 + b x + c = 0 then evaluate
a sin2 (A + B) + b sin (A + B) . cos (A + B) + c cos2 (A + B).
Sol.
b
c
; tan A . tan B =
a
a
tan A + tan B =
b
a = b
c a
1 ac
tan (A + B) =
Now
a b2
1
=
b2
( c a )2
1
=
E = cos2 (A + B) [a tan2 (A + B) + b tan (A + B) + c]
a )2
(c
b
2
(c
(c
b2
a)
2
c
b2 c
a)
2
=
a )2
(c
b
2
(c
b2
a)
2
c
a
a c
a
1
c
E=c
c
a )2
(c
a
c
Ex.14 Show that cos2A + cos2(A + B) + 2 cosA cos(180Â° + B) Â· cos(360Â° + A + B) is independent of A. Hence
find its value when B = 810Â°.
Sol. cos2 A + cos2 (A + B) â [2 cosA Â· cosB Â· cos (A + B)]
cos2 A + cos2(A + B) â [ {cos(A + B) + cos(A â B) } cos (A + B) ]
cos2 A + cos2 (A + B) â cos2(A + B) â (cos2 A â sin2 B)
= sin2 B which is independent of A now, sin2(810Â°) = sin2 (720Â° + 90Â°) = sin2 90Â° = 1 Ans.
Ex.15 Simplify: cos x Â· sin(y â z) + cos y Â· sin(z â x) + cos z Â· sin (x â y) where x, y, z R.
Sol. (1/2)[sin(y â z + x) + sin(y â z â x) + sin(z â x + y) + sin(z â x â y) + sin(x â y + z) + sin(x â y â z)] = 0
C.
MULTIPLE ANGLES AND SUB-MULTIPLE ANGLES
(a)
sin 2A = 2 sinA cosA ; sin = 2 sin
(b)
cos 2A = cosÂ²A
cos
= cosÂ²
2
2
cos
sinÂ²A = 2cosÂ²A 1 = 1
sinÂ²
2
= 2cosÂ²
2
1= 1
2
2 sinÂ²A ;
2sinÂ²
2 cosÂ²A = 1 + cos 2A , 2sinÂ²A = 1
cos 2A ;
2 cosÂ²
cos
2
= 1 + cos
, 2 sinÂ²
2
=1
2
.
.
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```
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