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.
Answer: 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.
Answer: 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°)`
Answer:
`(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°`
Answer: `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
Answer: (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
Answer: (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
Answer: (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
Answer: (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`