10th Maths 8.2
Chapter 8
Introduction to Trigonometry
NCERT Class 10th solution of Exercise 8.1
`text{Trigonometric Ratios of }``text{Some Specific Angles}`
Exercise 8.2
Q1. Evaluate the following:
i) `sin 60^circ cos 30^circ + sin 30^circ cos 60^circ`
Sol. :
`= sin 60^circ cos 30^circ + sin 30^circ cos 60^circ`
`= sqrt3/2 times sqrt3/2 + 1/2 times 1/2`
`= 3/4 + 1/4`
`= 4/4`
`= 1`
Answer:
`= 1`
ii) `2 tan^2 45^circ + cos^2 30^circ - sin^2 60circ`
Sol. :
`= 2 tan^2 45^circ + cos^2 30^circ - sin^2 60circ`
`= 2 times (1)^2 + (sqrt3/2)^2 - (sqrt3/2)^2`
`= 2 + 3/4 - 3/4`
` = 2`
Answer:
`= 2`
iii) `(cos 45^circ)/(sec 30^circ + cosec 30^circ)`
Sol. :
`= (cos 45^circ)/(sec 30^circ + cosec 30^circ)`
`= (1/sqrt2)/(2/sqrt3 + 2)`
`= (1/sqrt2)/((2 + 2sqrt3)/(sqrt3))`
`= 1/sqrt2 times sqrt3/(2 + 2sqrt3)`
`= sqrt3/(2sqrt2 + 2sqrt6)`
`= sqrt3/(2(sqrt6 + sqrt2)) times (sqrt6 - sqrt2)/(sqrt6 - sqrt2)`
`= (sqrt3(sqrt6 - sqrt2))/(2(6 - 2)`
`= (3sqrt2 - sqrt6)/8`
Answer:
`= (3sqrt2 - sqrt6)/8`
iv) `(sin 30^circ + tan 45^circ - cosec 60^circ)/(sec 30^circ + cos 60^circ + cot 45^circ)`
Sol. :
`= (sin 30^circ + tan 45^circ - cosec 60^circ)/(sec 30^circ + cos 60^circ + cot 45^circ)`
`= (1/2 +1 - 2/sqrt3)/(2/sqrt3 + 1/2 + 1)`
`= ((sqrt3 + 2sqrt3 - 4)/(2sqrt3))/((4 + sqrt3 + 2sqrt3)/(2sqrt3))`
`= (3sqrt3 - 4)/(3sqrt3 + 4)`
`= (3sqrt3 - 4)/(3sqrt3 + 4) times (3sqrt3 - 4)/(3sqrt3 - 4)`
`= (3sqrt3 - 4)^2/((3sqrt3 + 4)(3sqrt3 - 4))`
`= (27 + 16 - 24sqrt3)/(27 - 16)`
`= (43 - 24sqrt3)/11`
Answer:
`= (43 - 24sqrt3)/11`
v) `(5 cos^2 60^circ + 4 sec^2 30^circ - tan^2 45^circ)/(sin^2 30^circ + cos^2 30^circ)`
Sol. :
`= (5 cos^2 60^circ + 4 sec^2 30^circ - tan^2 45^circ)/(sin^2 30^circ + cos^2 30^circ)`
`= (5 times (1/2)^2 + 4 times (2/sqrt3)^2 - (1)^2)/((1/2)^2 + (sqrt3/2)^2`
`= (5/4 + 16/3 - 1)/(1/4 + 3/4)`
`= ((15 + 64 - 12)/12)/1`
`= (79 - 12)/12`
`= 67/12`
Answer:
`= 67/12`
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Q2. Choose the correct option and justify your choice:
i) `(2 tan 30^circ)/(1 + tan^2 30^circ) =`
A) `sin 60^circ` B) `cos 60^circ` C) `tan 60^circ` D) `sin 30^circ`
Sol. :
`= (2 tan 30^circ)/(1 + tan^2 30^circ)`
`= (2 times 1/sqrt3)/(1 + (1/sqrt3)^2`
`= (2/sqrt3)/(1 + 1/3)`
`= (2/sqrt3)/(4/3)`
`= 2/sqrt3 times 3/4`
`= sqrt3/2`
`= sin 60^circ`
Answer:
(A) `sin 60^circ`
ii) `(1 - tan^2 45^circ)/(1 + tan^2 45^circ) =`
A) `tan 90^circ` B) `1` C) `sin45^circ` D) `0`
Sol. :
`= (1 - tan^2 45^circ)/(1 + tan^2 45^circ)`
`= (1 - (1)^2)/(1 + (1)^2)`
`= (1 - 1)/(1 + 1)`
`= 0/2`
`= 0`
Answer:
(D) `0`
iii) `sin 2A = 2 sin A` is true when `A =`
A) `0^circ` B) `30^circ` C) `45^circ` D) `60^circ`
Sol. :
`sin 2A = 2 sin A`
`sin(2 times 0^circ) = 2 sin 0`
`sin 0 = 2 sin 0`
`0 = 2 times 0`
`= 0`
Answer:
(A) `0`
iv) `(2 tan 30^circ)/(1 - tan^2 30^circ) =`
A) `cos 60^circ` B) `sin 60^circ` C) `tan 60^circ` D) `sin 30^circ`
Sol. :
`= (2 tan 30^circ)/(1 - tan^2 30^circ)`
`= (2 times 1/sqrt3)/(1 - (1/sqrt3)^circ`
`= (2/sqrt3)/(1 - 1/3)`
`= (2/sqrt3)/(2/3)`
`= 2/sqrt3 times 3/2`
`= sqrt3`
`= tan 60^circ`
Answer:
(C) `tan 60^circ`
Q3. If `tan (A + B) = sqrt3` and `tan (A - B) = 1/sqrt3 ; ``0^circ lt A + B le 90^circ ; A lt B,` find `A` and `B.`
Sol. :
`tan(A + B) = sqrt3`
`tan(A + B) = tan 60^circ`
`A + B = 60^circ`_________(1)
and
`tan(A - B) = 1/sqrt3`
`tan(A - B) = tan 30^circ`
`A - B = 30^circ`_________(2)
We add equations (1) and (2)
`A + B + A - B = 60^circ + 30^circ`
`2A = 90^circ`
`A = 90^circ/2`
`A = 45^circ`
this value is put in equation (1)
`45^circ + B = 60^circ`
`B = 60^circ - 45^circ`
`B = 15^circ`
Answer:
`A = 45^circ, B = 15^circ`
Q4. State whether the following are true or false. Justify your answer:
i) `sin (A + B) = sin A + sin B.`
ii) The value of `sin theta` increases as `theta` increases.
iii) The value of `cos theta` increases as `theta` increases.
iv) `sin theta = cos theta` for all value of `theta.`
v) `cot A` is not defined for `A = 0^circ.`
Answer:
i) False ii) True iii) False iv) False v) False.
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