Class 10 Maths

# Trigonometry

## NCERT Exercise 8.4

### Part 1

Question – 1 - Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Solution: sin A can be expressed in terms of cot A, as follows:

text(sin A)=1/(text(cosec A)

=1/(sqrt(text(cosec)^2A))

=1/(sqrt(1+text(cot)^2A))

sec A can be expressed in terms of cot A, as follows:

text(sec A)=sqrt(text(sec)^2A)

=sqrt(1+text(tan)^2A)

=sqrt(1+(1)/(text(cot)^2A))

=sqrt((1+text(cot)^2A)/(text(cot)^2A)

=(sqrt(1+text(cot)^2A))/(text(cot)A)

tan A can be expressed in terms of cot A, as follows:

text(tan A) = 1/text(cotA)

Question – 2 - Write all the other trigonometric ratios of ∠A in terms of sec A.

Solution: Sin A can be written in terms of sec A as follows:

text(sin A)=sqrt(text(sin)^2A)

=sqrt(1-text(cos)^2A)

=sqrt(1-1/(text(sec)^2A))

=sqrt((text(sec)^2A-1)/(text(sec)^2A)

=sqrt(text(sec)^2A-1)/(text(sec)A)

cos A can be written in terms of sec A as follows:

text(cos A) = 1/text(sec A)

tan A can be written in terms of sec A as follows:

text(tan A)=sqrt(text(tan)^2A)

=sqrt(text(sec)^2A-1)

cosec A can be written in terms of sec A as follows:

text(cosec A)=1/(text(sin)A)

=1/sqrt(text(sin)^2A)

=1/sqrt(1-text(cos)^2A)

=1/sqrt(1-1/(text(sec)^2A))

=text(sec A)/sqrt(text(sec)^2A-1)

cot A can be written in terms of sec A as follows:

text(cot A)=1/text(tan A)

=1/sqrt(text(tan)^2A)

=1/sqrt(text(sec)^2A-1)

Question – 3 - Evaluate:

(i) (text(sin)^2\63°+text(sin)^2\27°)/(text(cos)^2\17°+text(cos)^2\73°)

Solution:

(text(sin)^2\63°+text(sin)^2\27°)/(text(cos)^2\17°+text(cos)^2\73°)

=(text(sin)^2\63°+text(sin)^2(90°-63°))/(text(cos)^2\17°+text(cos)^2(90°-17°))

=(text(sin)^2\63°+text(cos)^2\63°)/(text(cos)^2\17°+text(sin)^2\17°)=1

(ii) text(sin) 25°text(cos) 65°+text(cos) 25°text(sin) 65°

Solution: text(sin) 25°text(cos) 65°+ text(cos) 25°text(sin) 65°

= text(sin) 25° text(cos)(90°-25°) + text(cos) 25°text(sin) 65°

= text(sin)^2\25° + text(cos) 25° text(cos)(90°-25°)

= text(sin)^2\25° + text(cos)^2\25° = 1

Question – 4 - Choose the correct option. Justify your choice.

(i) 9 text(sec)^2\A – 9 text(tan)^2\A = ?
(A) 1 (B) 9 (C) 8 (D) 0

Solution: (B) is the correct option. Following is the explanation:

9 text(sec)^2\A – 9 text(tan)^2\A

= 9 (text(sec)^2\A – text(tan)^2\A)

= 9 xx 1 = 9

(ii) (1 + tan θ + sec θ) (1 + cot θ - cosec θ) = ?
(A) 0 (B) 1 (C) 2 (D) - 1

Solution: (C) is the correct option. Following is the explanation:

(1+text(tan)θ+text(sec)θ)(1+text(cot)θ-text(cosec)θ)

=(1+(text(sin)θ)/(text(cos)θ)+1/(text(cost)θ))(1+(text(cos)θ)/(text(sin)θ)-1/(text(sin)θ))

=((text(cos)θ+text(sin)θ+1)/(text(cos)θ))((text(sin)θ+text(cos)θ-1)/(text(sin)θ))

=(text(sin)θtext(cos)θ+text(sin)^2θ+text(sin)θ+text(sin)θtext(cos)θ
+text(cos)^2θ+text(cos)θ-text(cos)θ-text(sin)θ-1)
÷(text(sin)θtext(cos)θ)

(=(text(sin)^2θ+text(cos)^2θ+2text(sin)θtext(cos)θ+text(sin)θ
-text(sin)θ+text(cos)θ-text(cos)θ-1)
÷(text(sin)θtext(cos)θ)

=(1+2text(sin)θtext(cos)θ-1)/(text(sin)θtext(cos)θ)

=(2text(sin)θtext(cos)θ)/( text(sin)θtext(cos)θ)=2

(iii) (sec A + tan A)(1 – sin A) = ?
(A) sec A (B) sin A (C) cosec A (D) cos A

Solution: (D) is the correct option. Following is the explanation:

(text(sec)A+text(tan)A)(1-text(sin)A)

=(1/(text(cos)A)+(text(sin)A)/(text(cos)A))(1-text(sin)A)

=((1+text(sin)A)(1-text(sin)A))/(text(cos)A)

=(1-text(sin)^2A)/(text(cos)A)

=(text(cos)^2A)/(text(cos)A)=text(cos)A

(iv) (1+text(tan)^2A)/(1+text(cot)^2A)

(A) sec2 A, (B) - 1, (C) cot2 A, (D) tan2 A

Solution: (d) is the correct option. Following is the explanation:

(1+text(tan)^2A)/(1+text(cot)^2A)

=(1+(text(sin)^2A)/(text(cos)^2A))/(1+(text(cos)^2A)/(text(sin)^2A))

=((text(cos)^2A+text(sin)^2A)/(text(cos)^2A))/((text(sin)^2A+text(cos)^2A)/(text((sin)^2A))

=(1/(text(cos)^2A))/(1/(text(sin)^2A))

=(text(sin)^2A)/(text(cos)^2A)=text(tan)^2A