10th Maths 8.3
Chapter 8
Introduction to Trigonometry
NCERT Class 10th solution of Exercise 8.1
NCERT Class 10th solution of Exercise 8.4
Exercise 8.3
Q1. Evaluate:
i) `sin 18^circ/cos 72^circ`
Sol. :
`= sin 18^circ/cos 72^circ`
`= sin 18^circ/cos(90^circ - 18^circ)`
`= sin 18^circ/sin 18^circ`
`= 1`
Answer:
`= 1`.
ii) `tan 26^circ/cot 64^circ`
Sol. :
`= tan 26^circ/cot 64^circ`
`= tan 26^circ/cot(90^circ - 26^circ)`
`= tan 26^circ/tan 26^circ`
`= 1`
Answer:
`= 1`
iii) `cos 48^circ - sin 42^circ`
Sol. :
`= cos 48^circ - sin 42^circ`
`= cos(90^circ - 42^circ) - sin 42^circ`
`= sin 42^circ - sin 42^circ`
`= 0`
Answer:
`= 0`
iv) `cosec 31^circ - sec 59^circ`
Sol. :
`= cosec 31^circ - sec 59^circ`
`= cosec(90^circ - 59^circ) - sec 59^circ`
`= sec 59^circ - sec 59^circ`
`= 0`
Answer:
`= 0`
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Q2. Show that:
i) `tan 48^circ tan23^circ tan42^circ tan 67^circ = 1`
Sol. :
`L.H.S.`
`= tan 48^circ tan 23^circ tan 67^circ`
`= tan(90^circ - 42^circ) tan 23^circ tan 42^circ tan(90^circ - 23^circ)`
`= cot 42^circ tan 23^circ tan 42^circ cot 23^circ`
`= (cos 42^circ/sin 42^circ) times (sin 23^circ/cos 23^circ) times (cos 23^circ/sin 23^circ)`
`= 1`
`= R.H.S.`
`L.H.S. = R.H.S.`
Proved.
ii) `cos 38^circ cos 52^circ - sin 38^circ sin 52^circ =0`
Sol. :
`L.H.S.`
`= cos 38^circ cos 52^circ - sin 38^circ sin 52^circ`
`= cos 38^circ cos(90^circ - 38^circ) ``- sin 38^circ sin(90^circ - 38^circ)`
`= sin 38^circ cos38^circ - sin 38^circ cos 38^circ`
`= 0`
Answer:
`= 0`
Q3. If `tan 2A = cot(A - 18^circ),` where `2A` is an acute, angle find the value of `A.`
Sol. :
`tan 2A = cot(A - 18^circ)`
`cot(90^circ - 2A) = cot(A - 18^circ)`
`90^circ - 2A = A - 18^circ`
`90^circ + 18^circ =A + 2A`
`108^circ = 3A`
`A = 108^circ/3`
`A = 36^circ`
Answer:
`A = 36^circ`.
Q4. If `tan A = cot B,` prove that `A + B = 90^circ.`
Sol. :
`tan A = cot B`
`tan A = tan(90^circ - B)`
`A = 90^circ - B`
`A + B = 90^circ`
Proved.
Q5. If `sec 4A = cosec (A - 20^circ),` where `4A` is an acute angle, find the value of `A.`
Sol. :
`sec 4A = cosec(A - 20^circ)`
`cosec(90^circ - 4A) = cosec(A - 20^circ)`
`90^circ - 4A = A - 20^circ`
`4A + A = 90^circ + 20^circ`
`5A = 110^circ`
`A = 110^circ/5`
`A = 22^circ`
Answer:
`A = 22^circ`
Q6. If `A, B` and `C` are interior angles of a triangle `ABC,` then show that
`sin((A + B)/2) = cos A/2`
Sol. :
Sum of interior angles of triangles `ABC`
`A + B + C = 180^circ`
`B + C = 180^circ - A`
Divide by `2` both sides
`(B + C)/2 = (180^circ - A)/2`
`(B + C)/2 = (90^circ - A/2)`
`L.H.S.`
`= sin((B + C)/2)`
`= sin(90^circ - A/2)`
`= cos A/2`
`= R.H.S.`
`L.H.S. = R.H.S.`
Proved.
Q7. Express `sin 67^circ + cos 75^circ` in terms of trigonometric ratios of angles between `0^circ` and `45^circ.`
Sol. :
`= sin 67^circ + cos 75^circ`
`= sin(90^circ - 23^circ) + cos(90^circ - 15^circ)`
`= cos 23^circ + sin 15^circ`
Answer:
`= cos 23^circ + sin 15^circ`.
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