(1) If tan − tan B = and cot − cot = then prove that cot( − ) = 1 + 1 .
(2) Prove that cos 29+sin29 cos 29−sin29 = tan 74°.
(3) If cot = √3 and lies in fourth quadrant then find the value of : cos(− 2 ) sec(−) + cosec( 2 + ) sec( + ) + tan( 2 + ) cot( + ) + sin(− 2 ) cos(−) .
(4) Prove that sin( + ) sin( − ) = sin2 − sin2 . Hence prove that sin2 ( 8 + 2 ) − sin2 ( 8 − 2 ) = 1 √2 sin.
(5) Without using calculator, find the value of the following problems. (a) sin(300°) tan(330°) sec(420°) cot(135°) cos(210°) cosec(315°) (b) sec(220°) cos(580°) − tan(940°) cot(580°)
(6) Prove that sin 20° sin 40° sin 60° sin 80° = 3 16 .
(7) If tan = 1 2 ,tan = 1 5 ,tan = 1 8 then prove that + + = 4
(8) Prove that 3 = 3−tan3 1−3 tan2 .
(9) Prove that 3 − 3 = 2
Answers
Step-by-step explanation:
(1) We have,
tanA−tanB=x and, cotB−cotA=y
Now,
cotB−cotA=y
⇒
tanB
1
−
tanA
1
=y
⇒
tanAtanB
tanA−tanB
=y
⇒
tanAtanB
x
=y
⇒tanAtanB=
y
x
Therefore,
cot(A−B)=
tan(A−B)
1
=
tanA−tanB
1+tanAtanB
=
x
1+
y
x
=
xy
x+y
=
x
1
+
y
1
(2)search-icon-header
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Class 11
>>Maths
>>Trigonometric Functions
>>Trigonometric Functions of Sum and Difference of Two angles
>>Prove that: sin^2A + sin^2B + sin^2C = 2
Question
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Prove that: sin
2
A+sin
2
B+sin
2
C=2+2cosAcosBcosC.
Medium
Solution
verified
Verified by Toppr
On changing sin
2
A to 1−cos
2
A etc.
Proceeding directly as we need 2, we write sin
2
A=1−cos
2
A and
sin
2
B=1−cos
2
B
L.H.S.=2−cos
2
A−cos
2
B+sin
2
C
=2−(cos
2
A−sin
2
C)−cos
2
B
=2−cos(A+C)cos(A−C)−cos
2
B
=2+cosBcos(A−C)−cos
2
B
=2+cosB[cos(A−C)−cosB]
=2+cosB[cos(A−C)+cos(A+C)]
=2+cosB(2cosAcosC)
=2+2cosAcosBcosC.