Question 2 Choose the correct option and justify your choice.
(i) 2tan30 / (1+tan^2 30) =
(A). sin60°
(B). cos60°
(C). tan60°
(D). sin30°
(ii) (1-tan^2 45) / (1+tan^2 45)
(A). tan90°
(B). 1
(C). sin45°
(D). 0
(iii) sin2A = 2sinA is true when A =
(A). 0°
(B). 30°
(C). 45°
(D). 60°
(iv) 2tan30 / (1 - tan^2 30)
(A). cos60°
(B). sin60°
(C). tan60°
(D). sin30°
Class 10 - Math - Introduction to Trigonometry Page 187
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(i)
=>tan 30=1/√3
=>tan45=1
2tan30/(1+tan²45)
=(2×1/√3)\(1+1)
=(2/√3)/2
=2/√3×1/2
=√3
and tan60=√3
So,option (C) is ryt.
(ii)(1-tan²45)/(1+tan²45)
=(1-1)/(1+1)
=0/2
=0
So,option (D) is ryt.
(iii)Sin2A=2SinA
putting A=0
Then,
LHS
Sin2×0=Sin0=0
RHS
2SinA=2Sin0=2×0=0
Since,LHS=RHS
So,option (A) is ryt.
(iv)2tan30/(1-tan²30)
=2×1/√3/(1-(1/√3)²)
=2/√3/(1-1/3)
=2/√3/3-1/3
=2/√3/2/3
=√3
and tan60=√3
so,option(C) is ryt.
@Altaf
=>tan 30=1/√3
=>tan45=1
2tan30/(1+tan²45)
=(2×1/√3)\(1+1)
=(2/√3)/2
=2/√3×1/2
=√3
and tan60=√3
So,option (C) is ryt.
(ii)(1-tan²45)/(1+tan²45)
=(1-1)/(1+1)
=0/2
=0
So,option (D) is ryt.
(iii)Sin2A=2SinA
putting A=0
Then,
LHS
Sin2×0=Sin0=0
RHS
2SinA=2Sin0=2×0=0
Since,LHS=RHS
So,option (A) is ryt.
(iv)2tan30/(1-tan²30)
=2×1/√3/(1-(1/√3)²)
=2/√3/(1-1/3)
=2/√3/3-1/3
=2/√3/2/3
=√3
and tan60=√3
so,option(C) is ryt.
@Altaf
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