EXERCISE 8.4
Q 1, Ex 8.4 - Trigonometry - Chapter 8 - Maths Class 10th - NCERT
Page No 193:
Question 1:
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.Answer:
We know that,will always be positive as we are adding two positive quantities.
Therefore,
We know that,
However,
Therefore,
Also,
Q 2, Ex 8.4 - Trigonometry - Chapter 8 - Maths Class 10th - NCERT
Question 2:
Write all the other trigonometric ratios of ∠A in terms of sec A.Answer:
We know that,Also, sin2 A + cos2 A = 1
sin2 A = 1 − cos2 A
tan2A + 1 = sec2A
tan2A = sec2A − 1
Answer:
(i)(As sin2A + cos2A = 1)
= 1
(ii) sin25° cos65° + cos25° sin65°
= sin225° + cos225°
= 1 (As sin2A + cos2A = 1)
Q 4 (i &ii), Ex 8.4 - Trigonometry - Chapter 8 - Maths Class 10th - NCERT
Q 4 (iii & iv), Ex 8.4 - Trigonometry - Chapter 8 - Maths Class 10th - NCERT
Question 4:
Choose the correct option. Justify your choice.(i) 9 sec2 A − 9 tan2 A =
(A) 1
(B) 9
(C) 8
(D) 0
(ii) (1 + tan θ + sec θ) (1 + cot θ − cosec θ)
(A) 0
(B) 1
(C) 2
(D) −1
(iii) (secA + tanA) (1 − sinA) =
(A) secA
(B) sinA
(C) cosecA
(D) cosA
(iv)
(A) sec2 A
(B) −1
(C) cot2 A
(D) tan2 A
Answer:
(i) 9 sec2A − 9 tan2A= 9 (sec2A − tan2A)
= 9 (1) [As sec2 A − tan2 A = 1]
= 9
Hence, alternative (B) is correct.
(ii)
(1 + tan θ + sec θ) (1 + cot θ − cosec θ)
Hence, alternative (C) is correct.
(iii) (secA + tanA) (1 − sinA)
= cosA
Hence, alternative (D) is correct.
(iv)
Hence, alternative (D) is correct.
Q 5 (i & ii), Ex 8.4 - Trigonometry - Chapter 8 - Maths Class 10th - NCERT
Q 5 (iii), Ex 8.4 - Trigonometry - Chapter 8 - Maths Class 10th - NCERT
Q 5 (vi), Ex 8.4 - Trigonometry - Chapter 8 - Maths Class 10th - NCERT
Q 5 (viii), Ex 8.4 - Trigonometry - Chapter 8 - Maths Class 10th - NCERT
Q 5 (ix), Ex 8.4 - Trigonometry - Chapter 8 - Maths Class 10th - NCERT
Q 5 (x), Ex 8.4 - Trigonometry - Chapter 8 - Maths Class 10th - NCERT
Question 5:
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.Answer:
(i)(ii)
(iii)
= secθ cosec θ +
= R.H.S.
(iv)
= R.H.S
(v)
Using the identity cosec2 = 1 + cot2,
L.H.S =
= cosec A + cot A
= R.H.S
(vi)
(vii)
(viii)
(ix)
Hence, L.H.S = R.H.S
(x)
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