NCERT Exercise 8.4

1

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Solution

We know that $cot^2 A+ 1 = cosec^2 A$ ⇒ $cosec A = \sqrt{( 1+cot^2 A)}$
$sin A= \dfrac{1}{cosec A} = \dfrac{1}{\sqrt{(1+cot²A)}}$

$tan A = \dfrac{1}{cot A}$

$sec A = \sqrt{(1 + tan^2A)}$

$∴ sec A = \sqrt{\left(1 + \dfrac{1}{cot^2A}\right)}$ $= \sqrt{\left(\dfrac{cot^2A + 1}{cot^2A}\right)}$ $= \dfrac{\sqrt{1 + cot^2A}}{cot A}$

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2

Write all the other trigonometric ratios of ∠ A in terms of sec A.

Solution
We know that $sin^2 A + cos^2 A = 1$
$∴ sin^2 A = 1 − cos^2 A$
⇒ $sin A= \sqrt{1-\left ( \dfrac{1}{secA} \right )^{2}}$ $=\sqrt{\dfrac{sec^{2}A-1}{sec^{2}A}} = \dfrac{\sqrt{sec^{2}A-1}}{secA}$


$tan^2A + 1 = sec^2A$
$∴ tan^2A = sec^2A − 1$ $⇒ tan A = \sqrt{sec^{2}A-1}$

$cotA= \dfrac{cosA}{sinA}=\dfrac{\dfrac{1}{secA}}{\dfrac{\sqrt{sec^{2}A-1}}{secA}}$ $=\dfrac{1}{\sqrt{sec^{2}A-1}}$

$cosecA=\dfrac{1}{sinA}=\dfrac{secA}{\sqrt{sec^{2}A-1}}$

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3

Evaluate:

(i) $\dfrac{sin^{2}63^°+sin^{2}27°}{cos^{2}17°+cos^{2}73°}$

(ii) $sin25°cos65°+cos25°sin65°$

Solution

(i) $\dfrac{sin^{2}63^°+sin^{2}27°}{cos^{2}17°+cos^{2}73°}$

$= \dfrac{sin^{2}63^°+sin^{2}(90° - 63°)}{cos^{2}(90° - 73°)+cos^{2}73°}$

$= \dfrac{sin^{2}63^°+cos^{2}63°}{sin^{2}73°+cos^{2}73°}$ = 1



(ii) $sin25°cos65° + cos25°sin65°$

$= sin25°cos(90° - 65°) + cos25°sin(90° - 65°)$

$= sin^225° + cos^225°$ = 1

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4

Choose the correct option. Justify your choice.

(i) $9sec^{2}A - 9tan^{2}A =$

(A) 1
(B) 9
(C) 8
(D) 0

(ii) $(1+tanθ+secθ)(1+cotθ-cosecθ )=$

(A) 0
(B) 1
(C) 2
(D) –1

(iii)$(sec A + tan A) (1 – sin A) =$

(A) sec A
(B) sin A
(C) cosec A
(D) cos A

(iv) $\dfrac{1+tan^{2}A}{1+cot^{2}A}$

(A) $sec^{2}A$
(B) -1
(C) $cot^{2}A$
(D) $tan^{2}A$

Solution
(i) $9sec^{2}A-9tan^{2}A$
= 9 $(sec^2A − tan^2A)$
= 9 (1) = 9
[ ∵ sec2 A − tan2 A = 1]

Hence, option (B) is correct.

(ii) $(1 + tan θ + sec θ) (1 + cot θ − cosec θ)$
$=\left(1+\dfrac{sinθ}{cosθ}+\dfrac{1}{cosθ}\right) \left(1+\dfrac{cosθ}{sinθ} - \dfrac{1}{sinθ } \right )$
$=\left(\dfrac{cosθ + sinθ +1}{cosθ } \right ) \left ( \dfrac{sinθ +cosθ -1}{sinθ } \right )$ $\dfrac{\left ( sinθ +cosθ \right )^{2}-(1)^{2}}{sinθ cosθ }$ =2

Hence,option (C) is correct.

(iii) $(secA + tanA) (1 − sinA)$
$= \left( \dfrac{1}{cosA}+\dfrac{sinA}{cosA} \right)\left(1 - sinA\right)$ $=\left (\dfrac{1 + sinA}{cosA} \right )\left ( 1-sinA \right )$ $=\dfrac{1-sin^{2}A}{cosA}$=$\dfrac{cos^{2}}{cosA} = cosA$

Hence, option (D) is correct.

(iv) $\dfrac{1+tan^{2}A}{1+cot^{2}A}$
$\dfrac{1+\dfrac{sin^{2}A}{cos^{2}A}}{1+\dfrac{cos^{2}A}{sin^{2}A}}$ $=\dfrac{\dfrac{cos^{2}A+sin^{2}A}{cos^{2}A}}{\dfrac{sin^{2}A+cos^{2}A}{sin^{2}A}}$ $=\dfrac{\dfrac{1}{cos^{2}A}}{\dfrac{1}{sin^{2}A}}$ $=\dfrac{sin^{2}A}{cos^{2}A}= tan^{2}A$

Hence, option (D) is correct.

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5

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) $(cosecθ - cotθ)^{2}=\dfrac{1-cosθ }{1+cosθ}$

(ii) $\dfrac{cosA}{1+sinA} + \dfrac{1+sinA}{cosA}=2secA$

(iii) $\dfrac{tanθ}{1-cotθ} + \dfrac{cotθ}{1 - tanθ} = 1+secθ cosecθ$

[Hint : Write the expression in terms of sinθ and cosθ]

(iv) $\dfrac{1+secA}{secA}=\dfrac{sin^{2}A}{1-cosA}$

[Hint :Simplify LHS and RHS separately]

(v) $\dfrac{cosA-sinA+1}{cosA+sinA-1}=cosecA+cotA$, using the identity $cosec^{2}A=1+cot^{2}A.$

(vi) $\sqrt{\dfrac{1+sinA}{1-sinA}}=secA+tanA$

(vii) $\dfrac{sinθ -2sin^{3}θ }{2cos^{3}θ -cosθ }=tanθ$

(viii) $(sinA+cosecA)^{2}+(cosA+secA)^{2}$=$7+tan^{2}A+cot^{2}A$

(ix) $(cosecA-sinA)(secA-cosA)$=$\dfrac{1}{tanA+cotA}$

[Hint: Simplify LHS and RHS separately]

(x) $(\dfrac{1+tan^{2}A}{1+cot^{2}A})$=$(\dfrac{1-tanA}{1-cotA})^{2}=tan^{2}A$

Solution
(i) LHS = $(cosecθ - cotθ)^{2}$
$= \left(\dfrac{1}{sinθ} - \dfrac{cosθ}{sinθ}\right)^2$
$= \left(\dfrac{1 - cosθ}{sinθ}\right)^2$
$= \dfrac{(1 - cosθ)^2}{sin^2θ}$
$= \dfrac{(1 - cosθ)(1 - cosθ)}{(1 - cosθ)(1 + cosθ)}$
$= \dfrac{1 - cosθ}{1 + cosθ}$ = RHS

Hence proved.

(ii) LHS = $\dfrac{cosA}{1 + sinA} + \dfrac{1 + sinA}{cosA}$
$= \dfrac{cos^2A + sin^2A + 1 + 2sinA}{(1 + sinA)cosA}$
$= \dfrac{2 + 2sinA}{(1 + sinA)cosA}$
$= \dfrac{2(1 + sinA)}{(1 + sinA)cosA}$
$= \dfrac{2}{cosA}$
$= 2 secA$ = RHS

Hence proved.

(iii) LHS = $\dfrac{tanθ}{1-cotθ} + \dfrac{cotθ}{1 - tanθ}$

$= \dfrac{\dfrac{sinθ}{cosθ}}{1-\dfrac{cosθ}{sinθ}} + \dfrac{\dfrac{cosθ}{sinθ}}{1 - \dfrac{sinθ}{cosθ}}$

$= \dfrac{sinθsinθ}{cosθ(1-cosθ)} + \dfrac{cosθcosθ}{sinθ(1 - sinθ)}$

$= \dfrac{sin^3θ - cos^3θ}{sinθcosθ(sinθ - cosθ)}$

$= \dfrac{(sinθ - cosθ)(sin^2θ + cos^2θ + sinθcosθ)}{sinθcosθ(sinθ - cosθ)}$

$= \dfrac{(sin^2θ + cos^2θ + sinθcosθ)}{sinθcosθ}$

$= \dfrac{(1 + sinθcosθ)}{sinθcosθ}$

$= \dfrac{sinθcosθ}{sinθcosθ} + \dfrac{1}{cosθ} × \dfrac{1}{sinθ}$

$= 1 + secθcosecθ$ = RHS

Hence proved.


(iv) LHS = $\dfrac{1+secA}{secA}$ $= \dfrac{1+\dfrac{1}{cosA}}{\dfrac{1}{cosA}}$ $= 1 + cosA$

Multiplying and dividing by $(1 - cos A)$

$= \dfrac{(1 + cosA)(1 - cosA)}{1 - cosA}$

$= \dfrac{(1 - cos^2A)}{1 - cosA}$

$= \dfrac{sin^2A}{1 - cosA}$ = RHS

Hence proved


(v) LHS = $\dfrac{cosA-sinA+1}{cosA+sinA-1}$

Dividing numerator and denominator by $sinA$
$= \dfrac{\dfrac{cosA}{sinA} - \dfrac{sinA}{sinA} + \dfrac{1}{sinA}}{\dfrac{cosA}{sinA} + \dfrac{sinA}{sinA} - \dfrac{1}{sinA}}$ $= \dfrac{cotA - 1 + cosecA}{cotA + 1 - cosecA}$ $= \dfrac{cotA + cosecA - 1}{cotA - cosecA + 1}$

Using the identity $cosec^{2}A=1+cot^{2}A$ ⇒ $cosec^{2}A - cot^{2}A = 1$

$= \dfrac{cotA + cosecA - (cosec^{2}A - cot^{2}A)}{cotA - cosecA + 1}$

$= \dfrac{cotA + cosecA - (cosecA + cotA)(cosecA - cotA)}{cotA - cosecA + 1}$

$= \dfrac{(cotA + cosecA) (1 - (cosecA - cotA))}{cotA - cosecA + 1}$

$= \dfrac{(cotA + cosecA) (1 - cosecA + cotA)}{cotA - cosecA + 1}$

$= cosecA + cotA$ = RHS

Hence proved.


(vi) LHS = $\sqrt{\dfrac{1+sinA}{1-sinA}}$

$= \sqrt{\dfrac{(1+sinA)(1+sinA)}{(1-sinA)(1+sinA)}}$

$= \sqrt{\dfrac{(1+sinA)^2}{1-sin^2A}}$

$= \sqrt{\dfrac{(1+sinA)^2}{cos^2A}}$

$= \dfrac{1+sinA}{cosA}$

$= \dfrac{1}{cosA} + \dfrac{sinA}{cosA}$

$= secA + tanA$ = RHS

Hence proved.


(vii) LHS = $\dfrac{sinθ - 2sin^{3}θ }{2cos^{3}θ -cosθ }$
$=\dfrac{sinθ(1 - 2sin^{2}θ) }{cosθ(2cos^{2}θ - 1) }$
$=\dfrac{sinθ(sin^{2}θ + cos^{2}θ - 2sin^{2}θ) }{cosθ(2cos^{2}θ - sin^{2}θ - cos^{2}θ) }$
$=\dfrac{sinθ(cos^{2}θ - sin^{2}θ) }{cosθ(cos^{2}θ - sin^{2}θ)}$
$=\dfrac{sinθ}{cosθ}$
$=tanθ$ = RHS

Hence proved.


(viii) LHS = $(sinA+cosecA)^{2}+(cosA+secA)^{2}$
$= sin^2A + cosec^2A + 2 sinAcosecA + cos^2A + sec^2A + 2cosAsecA$
$= (sin^2A + cos^2A) + cosec^2A + 2 + sec^2A + 2$
$= 1 + (1 + cot^2A) + 2 + (1 + cot^2A) + 2$
$= 7 + tan^{2}A + cot^{2}A$ = RHS

Hence proved.


(ix) LHS = $(cosecA-sinA)(secA-cosA)$

$= \left(cosecA-\dfrac{1}{cosecA}\right)\left(secA-\dfrac{1}{secA}\right)$
$= \left(\dfrac{cosec^2A-1}{cosecA}\right)\left(\dfrac{sec^2A - 1}{secA}\right)$
$= \left(\dfrac{cot^2A}{cosecA}\right)\left(\dfrac{tan^2A}{secA}\right)$
$= \left(cot^2A sinA)\right)\left(tan^2AcosA)\right)$
$= sinAcosA$

RHS = $\dfrac{1}{tanA+cotA}$
$= \dfrac{1}{\dfrac{sinA}{cosA}+\dfrac{cosA}{sinA}}$

$= \dfrac{sinA cosA}{sin^2A + cos^2}$

$= sinAcosA$

Thus LHS = RHS

(x) LHS = $\left(\dfrac{1+tan^{2}A}{1+cot^{2}A}\right)$
$= \left(\dfrac{sec^{2}A}{cosec^{2}A}\right)$ $= \left(\dfrac{\dfrac{1}{cos^{2}A}}{\dfrac{1}{sin^{2}A}}\right)$ $= \left(\dfrac{sin^{2}A}{cos^{2}A}\right)$ $= tan^{2}A$

RHS = $(\dfrac{1-tanA}{1-cotA})^{2}$

$= \left(\dfrac{1-\dfrac{sinA}{cosA}}{1-\dfrac{cosA}{sinA}}\right)^{2}$ $= \left(\dfrac{\dfrac{cosA - sinA}{cosA}}{\dfrac{sinA - cosA}{sinA}}\right)^{2}$ $= \left(\dfrac{sinA}{cosA}\right)^{2}$ $= \left(tanA\right)^{2}$ $= tan^{2}A$

Thus LHS = RHS