Introduction to Trigonometry - Exercise 8.3

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Overview

Step-by-step NCERT solutions and practice problems for CBSE Class 10 Introduction to Trigonometry (Exercise 8.3), proving fundamental trigonometric identities ($sin^2 A + cos^2 A = 1, 1 + an^2 A = sec^2 A, 1 + cot^2 A = ext{cosec}^2 A$) with complete algebraic rigor.

Trigonometric Identities & Algebraic Proofs

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Key Identities Q1: Terms of cot A Q2: Terms of sec A Q3: MCQs Q4: Proving Identities
Three Fundamental Trigonometric Identities: $$ ext{1. } sin^2 A + cos^2 A = 1 implies sin^2 A = 1 - cos^2 A, quad cos^2 A = 1 - sin^2 A$$ $$ ext{2. } 1 + an^2 A = sec^2 A implies sec^2 A - an^2 A = 1, quad an^2 A = sec^2 A - 1$$ $$ ext{3. } 1 + cot^2 A = ext{cosec}^2 A implies ext{cosec}^2 A - cot^2 A = 1, quad cot^2 A = ext{cosec}^2 A - 1$$
Question 1 NCERT Ex 8.3
Express the trigonometric ratios $sin A$, $sec A$ and $ an A$ in terms of $cot A$.
Part (i): Express $ an A$ in Terms of $cot A$
Using the reciprocal identity: $$ an A = rac{1}{cot A}$$
Part (ii): Express $sec A$ in Terms of $cot A$
Using identity $sec^2 A = 1 + an^2 A$: $$sec^2 A = 1 + left( rac{1}{cot A} ight)^2 = 1 + rac{1}{cot^2 A} = rac{cot^2 A + 1}{cot^2 A}$$ Taking the square root: $$sec A = rac{sqrt{1 + cot^2 A}}{cot A}$$
Part (iii): Express $sin A$ in Terms of $cot A$
Using reciprocal identity $sin A = rac{1}{ ext{cosec } A}$ and identity $ ext{cosec}^2 A = 1 + cot^2 A$: $$ ext{cosec } A = sqrt{1 + cot^2 A}$$ $$sin A = rac{1}{sqrt{1 + cot^2 A}}$$
$ an A = rac{1}{cot A}$, $sec A = rac{sqrt{1 + cot^2 A}}{cot A}$, $sin A = rac{1}{sqrt{1 + cot^2 A}}$.
Question 2 NCERT Ex 8.3
Write all the other trigonometric ratios of $angle A$ in terms of $sec A$.
Expressing All 5 Ratios
1. $cos A$: $$cos A = rac{1}{sec A}$$ 2. $sin A$: $$sin^2 A = 1 - cos^2 A = 1 - rac{1}{sec^2 A} = rac{sec^2 A - 1}{sec^2 A} implies sin A = rac{sqrt{sec^2 A - 1}}{sec A}$$ 3. $ an A$: $$ an^2 A = sec^2 A - 1 implies an A = sqrt{sec^2 A - 1}$$ 4. $cot A$: $$cot A = rac{1}{ an A} = rac{1}{sqrt{sec^2 A - 1}}$$ 5. $ ext{cosec } A$: $$ ext{cosec } A = rac{1}{sin A} = rac{sec A}{sqrt{sec^2 A - 1}}$$
All ratios of $angle A$ are successfully represented in terms of $sec A$.
Question 3 NCERT Ex 8.3
Choose the correct option. Justify your choice:

(i) $9 sec^2 A - 9 an^2 A =$
(A) $1$    (B) $9$    (C) $8$    (D) $0$

(ii) $(1 + an heta + sec heta)(1 + cot heta - ext{cosec } heta) =$
(A) $0$    (B) $1$    (C) $2$    (D) $-1$

(iii) $(sec A + an A)(1 - sin A) =$
(A) $sec A$    (B) $sin A$    (C) $ ext{cosec } A$    (D) $cos A$

(iv) $ rac{1 + an^2 A}{1 + cot^2 A} =$
(A) $sec^2 A$    (B) $-1$    (C) $cot^2 A$    (D) $ an^2 A$
Part (i)
$$9 sec^2 A - 9 an^2 A = 9(sec^2 A - an^2 A)$$ Using identity $sec^2 A - an^2 A = 1$: $$= 9(1) = 9$$ Hence, Option (B) is correct.
Part (ii)
Convert into $sin heta$ and $cos heta$: $$= left(1 + rac{sin heta}{cos heta} + rac{1}{cos heta} ight)left(1 + rac{cos heta}{sin heta} - rac{1}{sin heta} ight)$$ $$= left( rac{cos heta + sin heta + 1}{cos heta} ight)left( rac{sin heta + cos heta - 1}{sin heta} ight)$$ $$= rac{[(sin heta + cos heta) + 1][(sin heta + cos heta) - 1]}{sin heta cos heta}$$ $$= rac{(sin heta + cos heta)^2 - 1^2}{sin heta cos heta} = rac{(sin^2 heta + cos^2 heta + 2sin heta cos heta) - 1}{sin heta cos heta}$$ $$= rac{1 + 2sin heta cos heta - 1}{sin heta cos heta} = rac{2sin heta cos heta}{sin heta cos heta} = 2$$ Hence, Option (C) is correct.
Part (iii)
$$(sec A + an A)(1 - sin A) = left( rac{1}{cos A} + rac{sin A}{cos A} ight)(1 - sin A)$$ $$= left( rac{1 + sin A}{cos A} ight)(1 - sin A) = rac{(1 + sin A)(1 - sin A)}{cos A} = rac{1 - sin^2 A}{cos A}$$ Since $1 - sin^2 A = cos^2 A$: $$= rac{cos^2 A}{cos A} = cos A$$ Hence, Option (D) is correct.
Part (iv)
$$ rac{1 + an^2 A}{1 + cot^2 A} = rac{sec^2 A}{ ext{cosec}^2 A} = rac{ rac{1}{cos^2 A}}{ rac{1}{sin^2 A}} = rac{sin^2 A}{cos^2 A} = an^2 A$$ Hence, Option (D) is correct.
Answers: (i) (B) $9$, (ii) (C) $2$, (iii) (D) $cos A$, (iv) (D) $ an^2 A$.
Question 4 NCERT Ex 8.3
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:

(i) $( ext{cosec } heta - cot heta)^2 = rac{1 - cos heta}{1 + cos heta}$
(ii) $ rac{cos A}{1 + sin A} + rac{1 + sin A}{cos A} = 2 sec A$
(iii) $ rac{ an heta}{1 - cot heta} + rac{cot heta}{1 - an heta} = 1 + sec heta ext{cosec } heta$
(iv) $ rac{1 + sec A}{sec A} = rac{sin^2 A}{1 - cos A}$
(v) $ rac{cos A - sin A + 1}{cos A + sin A - 1} = ext{cosec } A + cot A$
(vi) $sqrt{ rac{1 + sin A}{1 - sin A}} = sec A + an A$
(vii) $ rac{sin heta - 2sin^3 heta}{2cos^3 heta - cos heta} = an heta$
(viii) $(sin A + ext{cosec } A)^2 + (cos A + sec A)^2 = 7 + an^2 A + cot^2 A$
(ix) $( ext{cosec } A - sin A)(sec A - cos A) = rac{1}{ an A + cot A}$
(x) $left( rac{1 + an^2 A}{1 + cot^2 A} ight) = left( rac{1 - an A}{1 - cot A} ight)^2 = an^2 A$
Proof (i): $( ext{cosec } heta - cot heta)^2 = rac{1 - cos heta}{1 + cos heta}$
$$ ext{LHS} = left( rac{1}{sin heta} - rac{cos heta}{sin heta} ight)^2 = left( rac{1 - cos heta}{sin heta} ight)^2 = rac{(1 - cos heta)^2}{sin^2 heta}$$ Since $sin^2 heta = 1 - cos^2 heta = (1 - cos heta)(1 + cos heta)$: $$= rac{(1 - cos heta)^2}{(1 - cos heta)(1 + cos heta)} = rac{1 - cos heta}{1 + cos heta} = ext{RHS} quad ext{[Proved]}$$
Proof (ii): $ rac{cos A}{1 + sin A} + rac{1 + sin A}{cos A} = 2 sec A$
$$ ext{LHS} = rac{cos^2 A + (1 + sin A)^2}{cos A(1 + sin A)} = rac{cos^2 A + (1 + 2sin A + sin^2 A)}{cos A(1 + sin A)}$$ Using $cos^2 A + sin^2 A = 1$: $$= rac{1 + 1 + 2sin A}{cos A(1 + sin A)} = rac{2 + 2sin A}{cos A(1 + sin A)} = rac{2(1 + sin A)}{cos A(1 + sin A)} = rac{2}{cos A} = 2 sec A = ext{RHS} quad ext{[Proved]}$$
Proof (iii): $ rac{ an heta}{1 - cot heta} + rac{cot heta}{1 - an heta} = 1 + sec heta ext{cosec } heta$
Express in terms of $sin heta$ and $cos heta$: $$ rac{ an heta}{1 - rac{1}{ an heta}} + rac{cot heta}{1 - an heta} = rac{ an^2 heta}{ an heta - 1} - rac{1}{ an heta( an heta - 1)}$$ $$= rac{ an^3 heta - 1}{ an heta( an heta - 1)} = rac{( an heta - 1)( an^2 heta + an heta + 1)}{ an heta( an heta - 1)} = rac{ an^2 heta + an heta + 1}{ an heta}$$ $$= an heta + 1 + cot heta = 1 + rac{sin heta}{cos heta} + rac{cos heta}{sin heta} = 1 + rac{sin^2 heta + cos^2 heta}{sin heta cos heta}$$ $$= 1 + rac{1}{sin heta cos heta} = 1 + sec heta ext{cosec } heta = ext{RHS} quad ext{[Proved]}$$
Proof (iv): $ rac{1 + sec A}{sec A} = rac{sin^2 A}{1 - cos A}$
$$ ext{LHS} = rac{1 + rac{1}{cos A}}{ rac{1}{cos A}} = rac{ rac{cos A + 1}{cos A}}{ rac{1}{cos A}} = 1 + cos A$$ $$ ext{RHS} = rac{sin^2 A}{1 - cos A} = rac{1 - cos^2 A}{1 - cos A} = rac{(1 - cos A)(1 + cos A)}{1 - cos A} = 1 + cos A$$ Since $ ext{LHS} = ext{RHS} = 1 + cos A$, the identity is proved.
Proof (v): $ rac{cos A - sin A + 1}{cos A + sin A - 1} = ext{cosec } A + cot A$
Divide numerator and denominator by $sin A$: $$= rac{ rac{cos A}{sin A} - 1 + rac{1}{sin A}}{ rac{cos A}{sin A} + 1 - rac{1}{sin A}} = rac{cot A - 1 + ext{cosec } A}{cot A + 1 - ext{cosec } A} = rac{( ext{cosec } A + cot A) - 1}{cot A - ext{cosec } A + 1}$$ Substitute $1 = ext{cosec}^2 A - cot^2 A = ( ext{cosec } A - cot A)( ext{cosec } A + cot A)$ in numerator: $$= rac{( ext{cosec } A + cot A) - ( ext{cosec } A - cot A)( ext{cosec } A + cot A)}{cot A - ext{cosec } A + 1}$$ Factoring out $( ext{cosec } A + cot A)$: $$= rac{( ext{cosec } A + cot A)[1 - ( ext{cosec } A - cot A)]}{1 - ext{cosec } A + cot A} = rac{( ext{cosec } A + cot A)(1 - ext{cosec } A + cot A)}{1 - ext{cosec } A + cot A}$$ $$= ext{cosec } A + cot A = ext{RHS} quad ext{[Proved]}$$
Proof (vi): $sqrt{ rac{1 + sin A}{1 - sin A}} = sec A + an A$
Multiplying numerator and denominator inside the square root by $(1 + sin A)$: $$ ext{LHS} = sqrt{ rac{(1 + sin A)(1 + sin A)}{(1 - sin A)(1 + sin A)}} = sqrt{ rac{(1 + sin A)^2}{1 - sin^2 A}} = sqrt{ rac{(1 + sin A)^2}{cos^2 A}}$$ $$= rac{1 + sin A}{cos A} = rac{1}{cos A} + rac{sin A}{cos A} = sec A + an A = ext{RHS} quad ext{[Proved]}$$
Proof (vii): $ rac{sin heta - 2sin^3 heta}{2cos^3 heta - cos heta} = an heta$
Factoring out $sin heta$ from numerator and $cos heta$ from denominator: $$ ext{LHS} = rac{sin heta(1 - 2sin^2 heta)}{cos heta(2cos^2 heta - 1)} = an heta cdot rac{1 - 2sin^2 heta}{2cos^2 heta - 1}$$ Since $1 = sin^2 heta + cos^2 heta$: $$1 - 2sin^2 heta = sin^2 heta + cos^2 heta - 2sin^2 heta = cos^2 heta - sin^2 heta$$ $$2cos^2 heta - 1 = 2cos^2 heta - (sin^2 heta + cos^2 heta) = cos^2 heta - sin^2 heta$$ $$= an heta cdot rac{cos^2 heta - sin^2 heta}{cos^2 heta - sin^2 heta} = an heta cdot 1 = an heta = ext{RHS} quad ext{[Proved]}$$
Proof (viii): $(sin A + ext{cosec } A)^2 + (cos A + sec A)^2 = 7 + an^2 A + cot^2 A$
Expand both binomial squares: $$ ext{LHS} = (sin^2 A + 2sin A ext{cosec } A + ext{cosec}^2 A) + (cos^2 A + 2cos A sec A + sec^2 A)$$ Using $sin A ext{cosec } A = 1$ and $cos A sec A = 1$: $$= (sin^2 A + cos^2 A) + 2(1) + 2(1) + ext{cosec}^2 A + sec^2 A$$ $$= 1 + 2 + 2 + (1 + cot^2 A) + (1 + an^2 A) = 7 + an^2 A + cot^2 A = ext{RHS} quad ext{[Proved]}$$
Proof (ix): $( ext{cosec } A - sin A)(sec A - cos A) = rac{1}{ an A + cot A}$
$$ ext{LHS} = left( rac{1}{sin A} - sin A ight)left( rac{1}{cos A} - cos A ight) = left( rac{1 - sin^2 A}{sin A} ight)left( rac{1 - cos^2 A}{cos A} ight)$$ $$= rac{cos^2 A}{sin A} cdot rac{sin^2 A}{cos A} = sin A cos A$$ Now simplifying RHS: $$ ext{RHS} = rac{1}{ rac{sin A}{cos A} + rac{cos A}{sin A}} = rac{1}{ rac{sin^2 A + cos^2 A}{sin A cos A}} = rac{1}{ rac{1}{sin A cos A}} = sin A cos A$$ Since $ ext{LHS} = ext{RHS} = sin A cos A$, the identity is proved.
Proof (x): $left( rac{1 + an^2 A}{1 + cot^2 A} ight) = left( rac{1 - an A}{1 - cot A} ight)^2 = an^2 A$
Part 1: $$ rac{1 + an^2 A}{1 + cot^2 A} = rac{sec^2 A}{ ext{cosec}^2 A} = rac{ rac{1}{cos^2 A}}{ rac{1}{sin^2 A}} = rac{sin^2 A}{cos^2 A} = an^2 A$$ Part 2: $$left( rac{1 - an A}{1 - cot A} ight)^2 = left( rac{1 - an A}{1 - rac{1}{ an A}} ight)^2 = left( rac{1 - an A}{ rac{ an A - 1}{ an A}} ight)^2 = left(- an A ight)^2 = an^2 A$$ Both expressions equal $ an^2 A$. Hence proved.
All 10 trigonometric identities successfully proved with full mathematical rigor.
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