10th Maths 8.4
Chapter 8
Introduction to Trigonometry
NCERT Class 10th solution of Exercise 8.1
Exercise 8.4
Q1. Express the trigonometric ratios `sin A, sec A` and `tan A` in terms of `cot A`.
Sol. :
`sin A = 1/(cosec A) = 1/(sqrt (cosec^2 A)) = 1/sqrt(1 + cot^2 A)`
`sec A = 1/(cos A) = (1/sin A)/(cos A/sin A) = (cosec A)/(cot A)`
`sec A = sqrt(cosec^2 A)/(Cot A) = sqrt(1 + cot^2 A)/cot A`
`tan A = 1/cot A`
Answer:
`sin A = 1/sqrt(1 + cot^2 A)`, `sec A = sqrt(1 + cot^2 A)/cot A` and `tan A = 1/cot A`
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Q2. Write all the other trigonometric ratios of `angle A` in terms of `sec A`.
Sol. :
Let `triangle ABC` is right-angled at `angleB` and `sec A = x`
`sec A = x = (AC)/(AB)`
By Pythagoras theorem
`BC^2 = AC^2 - AB^2`
`BC^2 = x^2 - 1`
`BC = sqrt(x^2 -1)`
`sin A = (BC)/(AC) = sqrt(x^2 - 1)/x = sqrt(sec^2 - 1)/(sec A)`
`cos A = (AB)/(AC) = 1/x = 1/sec A`
`tan A = (BC)/(AB) = sqrt(x^2 - 1)/1 = sqrt(sec^2 A - 1)`
`cosec A = (AC)/(BC) = x/sqrt(x^2 - 1) = (sec A)/sqrt(sec A - 1)`
`cot A = (AB)/(BC) = 1/sqrt(x^2 - 1) = 1/sqrt(sec^2 A - 1)`
Answer:
`sin A = sqrt(sec^2 - 1)/(sec A)`
`cos A = 1/sec A`
`tan A = sqrt(sec^2 A - 1)`
`cosec A = (sec A)/sqrt(secA - 1)`
`cot A = 1/sqrt(sec^2 A - 1)`
Q3. Evaluate:
i) `(sin^2 63^circ + sin^2 27^circ)/(cos^2 17^circ + cos^2 73^circ)`
Sol. :
`= (sin^2(90^circ - 27^circ) + sin^2 27^circ)/(cos^2 17^circ + cos^2(90^circ - 17^circ)`
`= (cos^2 27^circ + sin^2 27^circ)/(cos^2 17^circ + sin^2 17^circ)`
`= 1/1`
`= 1`
Answer:
`= 1`
ii) `sin 25^circ cos 65^circ + cos 25^circ sin 65^circ`
Sol. :
`= sin 25^circ cos(90^circ - 25^circ) + cos 25^circ sin(90^circ - 25^circ)`
`= sin 25^circ sin 25^circ + cos 25^circ cos 25^circ`
`= sin^2 25^circ + cos^2 25^circ`
`= 1`
Answer:
`= 1`
Q4. Choose the correct option. Justify your choice:
i) `9 sec^2 A - 9 tan^2 A =`
A) `1` B) `9` C) `8` D) `0`
Sol. :
`= 9 sec^2 A - 9 tan^2 A`
`= 9(sec^2 - tan^2A)`
`= 9 times 1`
`= 9`
Answer:
`B) 9`.
ii) `(1 + tan theta + sec theta)(1 + cot theta - cosec theta) = `
A) `0` B) `1` C) `2` D) `-1`
Sol. :
`= (1 + tan theta + sec theta)(1 + cot theta - cosec theta)`
`= (1 + (sin theta)/(cos theta) + 1/(cos theta)) times (1 + (cos theta)/(sin theta) - 1/(sin theta))`
`= (cos theta + sin theta + 1)/(cos theta) times (sin theta + cos theta - 1)/(sin theta)`
`= ([(sin theta + cos theta) +1][(sin theta + cos theta) - 1])/(sin theta cos theta)`
`= ((sin theta + cos theta)^2 -1^2)/(sin theta cos theta)`
`= (sin^2 theta + cos^2 theta + 2 sin theta cos theta - 1)/(sin theta cos theta)`
`= (1 + 2 sin theta cos theta - 1)/(sin theta cos theta)`
`= (2sin theta cos theta)/(sin theta cos theta)`
`= 2`
Answer:
`D) 2`.
iii) `(sec A + tan A)(1 - sin A) =`
A) `sec A` B) sin A C) `cosec A` D) `cos A`
Sol. :
`= (sec A + tan A)(1 - sin A)`
`=(1/cos A + sin A/cos A)(1 - sin A)`
`= ((1 + sin A) (1 - sin A))/cos A`
`= (1 - sin^2 A)/cos A`
`= cos^2/cos A`
`= cos A`
Answer:
`D) cos A`
iv) `(1 + tan^2 A)/(1 + cot^2 A) =`
A) `sec^2 A` B) `- 1` C) `cot^2 A` D) `tan^2 A`
Sol. :
`= (1+ tan^2 A)/(1 + cot^2 A)`
`= (sec^2 A)/(cosec^2 A)`
`= 1/cos^2 A times sin^2 A/1`
`= (sin A/cos A)^2`
`= tan^2 A`
Answer:
`D) tan^2 A`.
Q5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
i) `(cosec theta - cot theta)^2 = (1 - cos theta)/(1 + cos theta)`
Sol. :
`LHS.`
`= (cosec theta - cot theta)^2`
`= (1/sin theta - cos theta/sin theta)^2`
`= ((1 - cos theta)/sin theta)^2`
`= (1 - cos theta)^2/sin^2 theta`
`= (1 - cos theta) ^2/(1 - cos^2 theta)`
`= (1 - cos theta)^2/((1 - cos theta)(1 + cos theta))`
`= (1 - cos theta)/(1 + cos theta)`
`= RHS.`
`LHS. = RHS.`
Proved.
ii) `cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A`
Sol. :
`= LHS.`
`= cos A/(1 + sin A) + (1 + sin A)/cos A`
`= (cos^2 A + (1 + sin A)^2)/(cosA(1 + sin A))`
`= (cos^2 A + 1 + sin^2 A + 2 sin A)/(cos A(1 + sin A))`
`= (sin^2 A + cos^2 A + 1 + 2 sin A)/(cos A(1 + sin A))`
`= (1 + 1 + 2 sin A)/(cos A(1 + sin A))`
`= (2 + 2 sin A)/(cos A(1 + sin A))`
`= (2(1 + sin A))/(cos A(1 + sin A))`
`= 2/cos A`
`= 2 sec A`
`= RHS.`
`LHS. = RHS.`
Proved.
iii) `tan theta/(1 - cot theta) + cot theta/(1 - tan theta) = 1 + sec theta cosec theta`
[Hint: Write the expression in terms of `sin theta` and `cos theta`]
Sol. :
`= LHS.`
`= tan theta/(1 - cot theta) + cot theta/(1 - tan theta)`
`= (sin theta/cos theta)/(1 - cos theta/sin theta) + (cos theta/sin theta)/(1 - sin theta/cos theta)`
`= sin^2 theta/(cos theta(sin theta - cos theta)) + cos^2 theta/(sin theta(cos theta - sin theta))`
`= (sin^3 theta - cos^3 theta)/(sin theta cos theta(sin theta - cos theta))`
`= (1 + sin theta cos theta)/(sin theta cos theta)`
`= 1/(sin theta cos theta) + 1`
`= cosec theta sec theta`
`= 1 + sec theta cosec theta`
`= RHS`
`LHS. = RHS.`
Proved.
iv) `(1 + sec A)/sec A = sin^2 A/(1 - cos A)` [Hint: Simplify `LHS` and `RHS` separately]
Sol. :
`= LHS.`
`= (1+ sec A)/sec A`
`= (1 + 1/cos A)/(1/cos A)`
`= ((cos A + 1)/cos A)/(1/cos A)`
`= (cos A + 1)/cos A times (cos A)/1`
`= (cos A + 1)/1`
`= (1+ cos A)/1 times (1 - cos A)/(1 - cos A)`
`= (1 - cos^2 A)/(1 - cos A)`
`= sin^2 A/(1 - cos A)`
`= RHS`
`LHS. = RHS.`
Proved.
v) `(cos A - sin A + 1)/(cosA + sinA - 1) = cosec A + cot A,` using the identity `cosec^2 A = 1 + cot A`
Sol. :
`= LHS.`
`= (cos A - sin A + 1)/(cos A + sin A - 1)`
`= (cot A - 1 + cosec A)/(cot A + 1 - cosec A)`
`= ((cot A - cosec A) - 1)/(cot A - cosec A + 1)`
`= ([(cot A + cosec A) - 1](cot A - cosec A))/([cot A - cosec A + 1](cot A - cosec A))`
`= ((cot^2 A - cosec^2 A) - (cot A - cosec A))/((cot A - cosec A +1)(cot A - cosec A))`
`= (- 1 - cot A + cosec A)/((cot A - cosec A +1)(cot A - cosec A))`
`= (cot A - cosec A + 1)/((cot A - cosec A + 1)(cosec A - cot A))`
`= 1/(cosec A - cot A)`
`= (cosec^2 A - cot^2 A)/(cosec A - cot A)`
`= ((cosec A - cot A)(cosec A + cot A))/(cosec A - cot A)`
`= cosec A + cot A`
`= RHS.`
`LHS. = RHS.`
Proved.
vi) `sqrt((1 + sin A)/(1 - sin A)) = sec A + tan A`
Sol. :
`= LHS.`
`= sqrt((1 + sin A)/(1 - sin A))`
`= sqrt((1 + sin A)/(1 - sin A)) times sqrt((1+ sin A)/(1 + sin A))`
`= sqrt((1 + sin A)^2/(1 - sin^2 A))`
`= sqrt((1 + sin A)^2/(cos^2 A))`
`= (1 + sin A)/cos A`
`= 1/cos A + sin A/cos A`
`= sec A + tan A`
`= RHS.`
`LHS. = RHS.`
Proved.
vii) `(sin theta - 2 sin^3 theta)/(2 cos^3 theta - cos theta) = tan theta`
Sol. :
`= LHS.`
`= (sin theta - 2sin^3 theta)/(2 cos^3 theta - cos theta)`
`= (sin theta(1 - 2 sin^2theta))/(cos theta (2 cos^2 theta - 1))`
`= (sin theta(1 - 2 sin^2 theta))/(cos theta[2(1 - sin^2 theta) - 1])`
`= (sin theta (1 - 2 sin^2 theta))/(cos theta(1 - 2 sin^2 theta)`
`= sin theta/cos theta`
`= tan theta`
`= LHS.`
`RHS. = LHS.`
Proved.
viii) `(sin A + cosec A)^2 + (cos A + sec A)^2 = 7 + tan^2 A + cot^2 A`
Sol. :
`= LHS.`
`= (sin A + cosec A)^2 + (cos A + sec A)^2`
`= sin^2 A + 2sin A cosec A + cosec^2 A + cos^2 A + 2cos A sec A + sec^2 A`
`= sin^2 A + cos^2 A + 2sin A times 1/sin A + cosec^2 A + sec^2 A + 2cos A times 1/cos A`
`= 1 + 2 + cosec^2 A + sec ^2 A + 2`
`= 1 + 2 + 2 + (1 + cot^2 A) + (1 + tan^2 A)`
`= 1 + 2 + 2 + 1 + 1 + cot^2 A + tan^2 A`
`= 7 + tan^2 A + cot^2 A`
`= RHS.`
`LHS. = RHS.`
ix) `(cosec A - sin A)(sec A - cos A) = 1/(tan A + cot A)`
[Hint: Simplify `LHS` and `RHS` separately]
Sol. :
`= LHS.`
`= (cosec A - sin A)(sec A - cos A)`
`= (1/sin A - sin A)(1/cos A - cos A)`
`= ((1- sin^2 A)/sin A)((1 - cos^2 A)/cos A)`
`= cos^2 A/sin A sin^2 A/cos A`
`= sin A cos A`
`= RHS.`
`= 1/(tan A + cot A)`
`= 1/(sin A/cos A + cos A/sin A)`
`= 1/((sin^2 A + cos^2 A)/(cos A sin A))`
`= 1/(1/(cosAsin A))`
`= sin A cos A`
`RHS. = LHS.`
Proved.
x) `((1 + tan^2 A)/(1 + cot^2 A)) = ((1 - tan A)/(1 - cot A))^2 = tan^2 A`
Sol. :
`= ((1 + tan^2 A)/(1 + cot^2 A))`
`= \frac{sec^2 A}{cosec^2 A}`
`= 1/cos^2 A times sin^2 A/1`
`= sin^2 A/cos^2 A`
`= tan^2 A`
`= ((1 - tan A)/(1 - cot A))^2`
`= ((1 - sin A/cos A)/(1 - cos A/sin A))^2`
`= [((cos A - sin A)/cos A)/((sin A - cos A)/sin A)]^2`
`= [(-((sin A - cos A))/cos A)/((sin A - cos A)/sin A)]^2`
`= (- sin A/cos A)^2`
`= (- tan A)^2`
`= tan^2`
`LHS. = RHS.`
Proved.
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