Question 1.1
Evaluate :
=
=
=1
Question 1.2
Evaluate:
=
=
= 1
Question 1.3
Evaluate:
=
=
= 1
Question 1.4
Evaluate:
=
=
= 1 + 1
= 2
Question 1.5
Evaluate: sin2 40° - cos2 50°
Sol:sin2 40° - cos2 50°
= sin2 (90° - 50°) - cos2 50°
= cos2 50° - cos2 50°
= 0
Question 1.6
Evaluate: sec2 18° - cosec2 72°
Sol:sec2 18° - cosec2 72°
= [sec (90° - 72°)]2 - cosec2 72°
= cosec2 72° - cosec2 72°
= 0
Question 1.7
Evaluate: sin 15° cos 15° - cos 75° sin 75°
Sol:sin 15° cos 15° - cos 75° sin 75°
= sin(90° - 75°) cos15° - cos 75° sin (90° -15°)
= cos 75° cos 15° - cos 75° cos 15°
= 0.
Question 1.8
Evaluate: sin 42° sin 48° - cos 42° cos 48°
Sol:sin 42° sin 48° - cos 42° cos 48°
= sin(90° - 48°)sin 48° - cos(90° - 48°)cos 48°
= cos 48° sin 48° - sin 48°cos 48°
= cos 48° sin 48° - cos 48° sin 48°
= 0
Question 2.1
Evaluate: sin (90° - A) sin A - cos (90° - A) cos A
Sol:sin (90° - A) sin A - cos (90° - A) cos A
= cos A sin A - sin A cos A
= 0
Question 2.2
Evaluate: sin2 35° - cos2 55°
Sol:sin2 35° - cos2 55°
= sin2 35° - [cos(90° - 35°)]2
= sin2 35° - sin2 35°
= 0
Question 2.3
Evaluate:
=
=
= 1 + 1 - 2
= 2 - 2
= 0
Question 2.4
Evaluate:
=
=
= 2 - 1
= 1
Question 2.5
Evaluate: cos2 25° - sin2 65° - tan2 45°
Sol:cos2 25° - sin2 65° - tan2 45°
= [cos(90° - 65°)]2 - sin2 65° - (tan 45°)2
= sin2 65° - sin2 65° - (1)2
= 0 - 1
= - 1.
Question 2.6
Evaluate:
=
=
=
= 1 + 1 - 1
= 1
Question 3.1
Show that: tan 10° tan 15° tan 75° tan 80° = 1.
Sol:L.H.S.
= tan 10° tan 15° tan 75° tan 80°
= tan (90° - 80°) tan (90° - 75°) tan 75° tan 80°
= cot 80° cot 75 ° tan 75° tan 80°
= (cot 80° tan 80°)(cot 75° tan 75°)
= (1)(1)
= 1
= R.H.S.
Question 3.2
Show that: sin 42° sec 48° + cos 42° cosec 48° = 2.
Sol:L.H.S.
= sin 42° sec 48° + cos 42° cosec 48°
=
=
= 1 + 1
= 2
= R.H.S.
Question 4.1
Express the following in term of angles between 0°and 45°:
sin 59°+ tan 63°
sin 59° + tan 63°
= sin(90 - 31)° + tan(90 - 27)°
= cos 31° + cot 27°.
Question 4.2
Express the following in term of angles between 0°and 45°:
cosec 68° + cot 72°
cosec 68° + cot 72°
= cosec(90 - 22)° + cot(90 - 18)°
(∵ cosec(90 - θ) = secθ and cot(90 - θ) = tanθ)
= sec 22° + tan 18°
Question 4.3
Express the following in term of angles between 0°and 45°:
cos 74°+ sec 67°
cos 74°+ sec 67°
= cos(90 - 16)° + sec(90 - 23)°
= sin 16° + cosec 23°
Question 5.1
For triangle ABC, show that:
We know that for a triangle ΔABC
∠A + ∠B + ∠C = 180°
Question 5.2
For triangle ABC, show that:
We know that for a triangle ΔABC
∠A + ∠B + ∠C = 180°
∠B + ∠C = 180° - ∠A
Question 6.1
Evaluate:
=
=
= 3 - 1
= 2
Question 6.2
Evaluate: 3 cos 80° cosec 10°+ 2 sin 59° sec 31°.
Sol:3 cos 80° cosec 10°+ 2 sin 59° sec 31°
= 3 cos (90° - 10°) cosec 10° + 2sin (90° - 31) sec 31°
= 3 sin 10° cosec 10° + 2 cos 31° sec 31°
= 3 x 1 + 2 x 1
= 3 + 2
= 5
Question 6.3
Evaluate:
=
=
= 1 + 1
= 2
Question 6.4
Evaluate: tan (55° - A) - cot (35° + A)
Sol:tan (55° - A) - cot (35° + A)
= tan [90° - (35° + A)] - cot (35° + A)
= cot (35° + A) - cot (35° + A)
= 0.
Question 6.5
Evaluate: cosec (65° + A) - sec (25° - A)
Sol:cosec (65° + A) - sec (25° - A)
= cosec [90° - (25° - A)] - sec (25° - A)
= sec (25° - A) - sec (25° - A)
= 0.
Question 6.6
Evaluate:
=
=
= 2 - 1 - 1
= 0.
Question 6.7
Evaluate:
=
=
= 1 - 2
= -1
=
=
= 1 - 2
= -1
Question 6.8
Evaluate:
=
=
= 1 + 1 - 2
= 0
Question 6.9
Evaluate: 14 sin 30°+ 6 cos 60°- 5 tan 45°.
Sol:14 sin 30°+ 6 cos 60°- 5 tan 45°
=
= 7 + 3 - 5
= 5.
Question 7
A triangle ABC is right-angled at B; find the value of
Since Δ ABC is a right angled triangle, right angled at B,
A + C = 90°
∴
=
=
=
= 1 - 1
= 0
Question 8.1
In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
sin (90° - 3A).cosec 42° = 1.
Sin (90° - 3A). cosec 42° = 1
⇒ sin (90° - 3A) =
⇒ cos 3A =
⇒ cos 3A =
⇒ cos 3A = cos 48°
⇒ 3A = 48°
⇒ A = 16°.
Question 8.2
In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
cos(90° - A).sec 77° = 1
cos (90° - 3A). sec 77° = 1
⇒ cos(90° - 3A) =
⇒ sin 3A =
⇒ sin 3A =
⇒ sin 3A - sin 12°
⇒ 3A = 12°
⇒ A = 3°
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