1. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = 12/5 for some value of angle A.
(iii)cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = 4/3 for some angle θ.
2.Choose the correct option and justify your choice :
(i) 2tan 30°/1+tan230° = (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
(ii) 1-tan245°/1+tan245° = (A) tan 90° (B) 1 (C) sin 45° (D) 0
(iii) sin 2A = 2 sin A is true when A = (A) 0° (B) 30° (C) 45° (D) 60°
(iv) 2tan30°/1-tan230° = (A) cos 60° (B) sin (C) tan 60° (D) sin 30°
Give answers
Answers
Answer:
False
∠B=90
o
, AB=3,BC=4,AC=5
tanA=
3
4
>d. ...(AC
2
=AB
2
+BC
2
)
(ii) True
AC
2
=AB
2
+BC
2
(12x)
2
=(5x)
2
+BC
2
,
BC=
119x
(it is possible) Pythagoras theorem
(iii) False
cosA is cosine A, cscA is cosecant
(iv) False
cotA = cotangent of ∠A not product of cot and A.
(v) False
sinθ=
hypotenuse
perpendicular
sinθ>1;
3
4
>1 [∵ hypotenuse > base > perpendicular]
sinθ will always less than 1.
PLEASE MAKE AS BRAINLIEST
q1
(I) The value of tan A is always less than 1.
Answer: False
Proof: In ΔMNC in which ∠N = 90∘,
MN = 3, NC = 4 and MC = 5
Value of tan M = 4/3 which is greater than.
The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as it will follow the Pythagoras theorem.
MC2=MN2+NC2
52=32+42
25=9+16
25 = 25
(ii) sec A = 12/5 for some value of angle A
Answer: True
Justification: Let a ΔMNC in which ∠N = 90º,
MC=12k and MB=5k, where k is a positive real number.
By Pythagoras theorem we get,
MC2=MN2+NC2
(12k)2=(5k)2+NC2
NC2+25k2=144k2
NC2=119k2
Such a triangle is possible as it will follow the Pythagoras theorem.
(iii) cos A is the abbreviation used for the cosecant of angle A.
Answer: False
Justification: Abbreviation used for cosecant of angle M is cosec M. cos M is the abbreviation used for cosine of angle M.
(iv) cot A is the product of cot and A.
Answer: False
Justification: cot M is not the product of cot and M. It is the cotangent of ∠M.
(v) sin θ = 4/3 for some angle θ.
Answer: False
Justification: sin θ = Height/Hypotenuse
We know that in a right angled triangle, Hypotenuse is the longest side.
∴ sin θ will always less than 1 and it can never be 4/3 for any value of θ.
q2
Solution:
(i) (A) is correct.
Substitute the of tan 30° in the given equation
tan 30° = 1/√3
2tan 30°/1+tan230° = 2(1/√3)/1+(1/√3)2
= (2/√3)/(1+1/3) = (2/√3)/(4/3)
= 6/4√3 = √3/2 = sin 60°
The obtained solution is equivalent to the trigonometric ratio sin 60°
(ii) (D) is correct.
Substitute the of tan 45° in the given equation
tan 45° = 1
1-tan245°/1+tan245° = (1-12)/(1+12)
= 0/2 = 0
The solution of the above equation is 0.
(iii) (A) is correct.
To find the value of A, substitute the degree given in the options one by one
sin 2A = 2 sin A is true when A = 0°
As sin 2A = sin 0° = 0
2 sin A = 2 sin 0° = 2 × 0 = 0
or,
Apply the sin 2A formula, to find the degree value
sin 2A = 2sin A cos A
⇒2sin A cos A = 2 sin A
⇒ 2cos A = 2 ⇒ cos A = 1
Now, we have to check, to get the solution as 1, which degree value has to be applied.
When 0 degree is applied to cos value, i.e., cos 0 =1
Therefore, ⇒ A = 0°
(iv) (C) is correct.
Substitute the of tan 30° in the given equation
tan 30° = 1/√3
2tan30°/1-tan230° = 2(1/√3)/1-(1/√3)2
= (2/√3)/(1-1/3) = (2/√3)/(2/3) = √3 = tan 60°
The value of the given equation is equivalent to tan 60°.
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