TS Inter 1st Year Maths 1A Ratios up to Transformations Important Questions Short Answer Type

Students must practice these Maths 1A Important Questions TS Inter 1st Year Maths 1A Ratios up to Transformations Important Questions Short Answer Type to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Ratios up to Transformations Important Questions Short Answer Type

Question 1.
Prove that \(\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}=\frac{1+\sin \theta}{\cos \theta}\). [Mar. ’14]
Answer:

Question 2.
If A + B = \(\frac{\pi}{4}\), then prove that i) (1 + tan A) (1 + tan B) = 2. ii) (cot A – 1) (cot B -1) = 2. [Mar. 61′(TS), ’07 Ma
Answer:
(i) Given that A + B = 45°
⇒ tan (A + B) = tan 45°
⇒ \(\frac{\tan A+\tan B}{1-\tan A \tan B}\) = 1
⇒ tan A + tan B = 1 – tan A . tan B
⇒ tan A + tan B + tan A tan B = 1

L.H.S = (1 + tan A) (1 + tan B)
= 1 + tan B + tan A + tan A tan B
= 1 + (tan A + tan B + tan A tan B)
= 1 + 1 = 2 = RHS.
∴ (1 + tan A) (1 + tan B) = 2

(ii) Given that A + B = 45°
⇒ cot (A + B) = cot 45°
⇒ \(\frac{\cot A \cot B-1}{\cot B+\cot A}\) = 1
⇒ cot A. cot B – 1 = cot A + cot B
⇒ cot A cot B – cot A – cot B = 1
L.H.S = (cot A – 1) (cot B – 1)
= cot A cot B – cot A – cot B + 1 = (cot A cot B – cot A – cot B) + 1 = 1 + 1=2 = RHS.
∴ (cot A – 1) (cot B – 1) = 2

Question 3.
Prove that sin²θ + sin²(θ + \(\frac{\pi}{3}\)) + sin²(θ – \(\frac{\pi}{3}\)) = \(\frac{\pi}{3}\)
Answer:

Question 4.
If A, B, C are the angles of a triangle and if none of them is equal to \(\frac{\pi}{2}\) then prove that tan A + tan B + tan C = tan A tan B tan C.
Sol. If A, B, C are the angles of a triangle then
A + B + C = 180°
⇒ A + B = 180 0 – C
⇒ tan (A + B) = tan (180° – C)
⇒ \(\frac{\tan A+\tan B}{1-\tan A \tan B}\) = – tan C
⇒ tan A + tan B = – tan C (1 – tan A tan B)
⇒ tan A + tan B = – tan C + tan A tan B tan C
⇒ tan A + tan B + tan C = tan A tan B tan C

Question 5.
If 0 < A < B < \(\frac{\pi}{4}\) and sin(A +B) = \(\frac{24}{25}\) and cos(A – B) = \(\frac{4}{5}\), then find the value of tan 2a.
Answer:
0 < A < \(\frac{\pi}{4}\) and 0 < B < \(\frac{\pi}{4}\) ⇒ 0 < A < B < \(\frac{\pi}{4}\)
∴ (A + B) lies in first quadrant.

Question 6.
If tan α – tan β = m and cot α – cot β = n, then prove that cot(α – β) = \(\frac{1}{m}-\frac{1}{n}\)
Answer:
We have tan α – tan β = m

Question 7.
If \(\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=\frac{a+b}{a-b}\) then prove that a tan β = b
Answer:
Given \(\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=\frac{a+b}{a-b}\)
By using componendo and dividendo, we get

Question 8.
If A + B + C = \(\frac{\pi}{2}\) and and if none of A, B, Cs an odd multiple of then prove that cot A + cot B + cot C = cot A cot B cot C.
Answer:
Given A + B + C = \(\frac{\pi}{2}\) ⇒ A + B = \(\frac{\pi}{2}\) – C
∴ cot(A + B) = c(\(\frac{\pi}{2}\) – C) = tanC = \(\frac{1}{\cot C}\)
⇒ \(\frac{\cot A \cot B-1}{\cot B+\cot A}=\frac{1}{\cot C}\)
⇒ cot A + cot B + cot C = cot A cot B cot C

Question 9.
Prove that sin 18° = \(\frac{\sqrt{5}-1}{4}\). [May ’10]
Answer:
Let A = 18°
5A = 90°
2A + 3A = 90°
2A = 90° – 3A
sin 2A = sin (90° – 3A)
sin 2A = cos 3A
2 sin A cos A = 4 cos³A – 3 cos A
2 sin A cos A = cos A (4 cos²A – 3)
2 sin A = 4 cos²A – 3
2 sin A = 4 (1 – sin²A) – 3
2 sin A = 4 – 4 sin²A – 3
2 sin A = 1 – 4 sin²A
4 sin²A + 2 sin A – 1 = 0
This is a quadratic equation in sin A.
Here a = 4,b = 2, c = -1

Since A = 18°, it lies in first quadrant and hence sin A > 0.
sin A = \(\frac{-1+\sqrt{5}}{4}\) ⇒ sin A = \(\frac{\sqrt{5}-1}{4}\)
∴ sin 18° = \(\frac{\sqrt{5}-1}{4}\)

Question 10.
If A is not an integral multiple of \(\frac{\pi}{2}\), prove that
(i) tan A + cot A = 2 cosec 2A,
(ii) cot A – tan A = 2 cot 2A. [B.P] [Mar. ’18(AP)]
Answer:

Question 11.
If θ is not an integral multiple of \(\frac{\pi}{2}\), prove that tan θ + 2 tan 2θ + 4 tan 4θ + 8 cot 8θ = cot θ. [Mar. ’19(AP); May ’01]
Answer:
We know that cot A – tan A = 2 cot 2A

cot A – tan A = 2 cot 2A
tan A = cot A – 2 cot 2A …………………(1)
Put A = 0 in (1)
tan θ = cot θ – 2 cot 2θ

Put A = 20 in (1)
tan 2θ = cot 2θ – 2 cot 4θ

Put A = 40 in (1)
tan 4θ = cot 4θ – 2 cot 8θ

L.H.S = tan θ + 2 tan 2θ + 4 tan 4θ + 8 cot 8θ = (cot θ – 2 cot 2θ) + 2 (cot 2θ – 2 cot 4θ) + 4 (cot 4θ – 2 cot 8θ) + 8 cot 8θ
= cot θ – 2 cot 2θ + 2 cot 2θ – 4 cot 4θ + 4 cot 4θ – 8 cot 8θ + 8 cot 8θ = cot θ = R.H.S
∴ tan θ + 2 tan 2θ + 4 tan 4θ + 8 cot 8θ = cot θ

Question 12.
Prove that sin A. sin(\(\frac{\pi}{3}\) + A) sin (\(\frac{\pi}{3}\) – A) = \(\frac{1}{4}\) sin 3A and hence deduce that sin 20°. sin 40°. sin 60°. sin 80° = \(\frac{3}{16}\). [May ’97, ’93]
Answer:

Let A = 20° ⇒ sin 20°. sin (60° + 20°) sin (60° – 20°)
= \(\frac{1}{4}\) sin 3 (20°)
⇒ sin 20° . sin 80°. sin 40° = \(\frac{1}{4}\) sin 60°

Multiply with sin 60° on both sides then we get
sin 20° . sin 40° . sin 60° . sin 80°
= \(\frac{1}{4}\)sin 60 – \(\frac{1}{4} \cdot\left(\frac{\sqrt{3}}{2}\right)^2=\frac{1}{4} \cdot \frac{3}{4}=\frac{3}{16}\)

Question 13.
Prove that sin⁴\(\frac{\pi}{8}\) + sin⁴\(\frac{5 \pi}{8}\) + sin⁴\(\frac{5 \pi}{8}\) + sin⁴\(\frac{7 \pi}{8}=\frac{3}{2}\). [Mar ’13; Mar. ’00]
Answer:

Question 14.
Show that sin A = \(\frac{\sin 3 A}{1+2 \cos 2 A}\). Hence find the value of sin 15°.
Answer:

Prove that tan α = \(\frac{\sin 2 \alpha}{1+\cos 2 \alpha}\) and hence deduce the values of tan 15° and tan 22\(\frac{1}{2}\)°
Answer:
2 – √3, √2 – 1

Question 15.
Prove that tan 9° – tan 27° – cot 27° + cot 9° = 4.
Answer:
We have tan A + cot A
= \(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}=\frac{1}{\sin A \cos A}\) = 2 cosec 2A
Put A = 9°, we have tan 9° + cot 9° = 2 cosec 18°
Put A = 27°, we have tan 27° + cot 27° = 2 cosec 54°
L.H.S = tan 9° – tan 27° + cot 9° – cot 27° = 2 (cosec 17° – cosec 54°)

Question 16.
Prove that cos²\(\frac{\pi}{8}\) + cos²\(\frac{3 \pi}{8}\) + cos²\(\frac{5 \pi}{8}\) + cos²\(\frac{7 \pi}{8}\) = 2.
Answer:

Question 17.
Prove that sin\(\frac{\pi}{5}\). sin\(\frac{2 \pi}{8}\).sin\(\frac{3 \pi}{8}\).sin\(\frac{4 \pi}{5}=\frac{5}{16}\). [Mar. ’13]
Answer:
L.H.S = sin\(\frac{\pi}{5}\). sin\(\frac{2 \pi}{8}\).sin\(\frac{3 \pi}{8}\).sin\(\frac{4 \pi}{5}\)
= sin 36° . sin 72°. sin 108°. sin 144°
= sin 36° . sin 72°. sin (90° + 18°) sin (180° – 36°)
= sin 36° . sin (90° -18°). sin (90° +18°). sin (180° – 36°)
= sin 36° . cos 18°. cos 18°. sin 36°
= sin²36° . cos²18°
= \(\left(\frac{10-2 \sqrt{5}}{16}\right)\left(\frac{10+2 \sqrt{5}}{16}\right)\)
= \(\frac{100-20}{256}=\frac{80}{256}=\frac{5}{16}\)

Question 18.
Prove that \(\frac{1-\sec 8 \alpha}{1-\sec 4 \alpha}=\frac{\tan 8 \alpha}{\tan 2 \alpha}\).
Answer:

Question 19.
Prove that [Mar. ’19 (AP) & TS]
(1 + cos\(\frac{\pi}{10}\))(1 + cos\(\frac{3 \pi}{8}\))(1 + cos \(\frac{7 \pi}{8}\))(1 + cos\(\frac{9 \pi}{8}\)) = \(\frac{1}{16}\)
Answer:
LHS = (1 + cos\(\frac{\pi}{10}\))(1 + cos\(\frac{3 \pi}{8}\))(1 + cos \(\frac{7 \pi}{8}\))(1 + cos\(\frac{9 \pi}{8}\)
(1 + cosl8) (1 + cos 54) (1 + cos 126) (1 + cos 162)
= (1 + cos 18) (1 + sin 36) (1 – sin 36) (1 – cos 18)
= (1 + cos 18) (1 – cos 18) (1 + sin 36) (1 – sin 36)
= (1 – cos218°) (1 – sin2 36°)
= sin²18° cos²36° = \(\left(\frac{\sqrt{5}-1}{4}\right)^2\left(\frac{\sqrt{5}+1}{4}\right)^2\)
= \(\frac{(5-1)^2}{256}=\frac{16}{256}=\frac{1}{16}\)

Question 20.
If A is not an integral multiple of t, prove that cos A. cos 2A. cos 4A . cos 8A = \(\frac{\sin 16 A}{16 \sin A}\) and hence deduce that cos\(\frac{2 \pi}{15}\).cos\(\frac{4 \pi}{15}\).cos\(\frac{8 \pi}{15}\).cos\(\frac{16 \pi}{15}=\frac{1}{16}\).
Answer:
L.H.S = cos A. cos 2A. cos 4A . cos 8A

Question 21.
If \(\frac{\sin (\alpha+\beta)}{\cos (\alpha-\beta)}=\frac{1-m}{1+m}\), then prove that tan(\(\frac{\pi}{4}\) – α) = m tan(\(\frac{\pi}{4}\) + β). [Mar. ’95, ’83, ’82, ’81]
Answer:
Given \(\frac{\sin (\alpha+\beta)}{\cos (\alpha-\beta)}=\frac{1-m}{1+m}\)
By using componendo and dividendo we get

Question 22.
If sec (θ + α) + sec (θ – α) = 2 sec θ and cos α ≠ 1, then show that cos θ = ± √2 cos \(\frac{α}{2}\).
Answer:
Given sec (θ + α) + sec (θ – α) = 2 sec θ

⇒ cos²θ cos α = cos²θ – sin²α
⇒ sin²α = cos²θ – cos²θ cos α
⇒ sin²α = cos²θ (1 – cos α)
⇒ 1 – cos²α = cos²θ (1 – cos α)
⇒ (1 – cos α) (1 + cos α) = cos²θ (1 – cos α)
⇒ cos²θ = 1 cos α ⇒ cos²θ = 2 cos²\(\frac{\alpha}{2}\)
⇒ cos θ = ± √2 cos \(\frac{α}{2}\)

Question 23.
If m sin B = n sin (2A + B), then prove that (m + n)tan A = (m – n)tan (A + B). [Apr. ’85]
Answer:
Given m sin B = n sin (2A + B)
⇒ \(\frac{m}{n}=\frac{\sin (2 A+B)}{\sin B}\)
By using componendo and dividendo we get