Rs Aggarwal 2020 2021 Solutions for Class 10 Maths Chapter 13 Trigonometric Identities are provided here with simple step-by-step explanations. These solutions for Trigonometric Identities are extremely popular among Class 10 students for Maths Trigonometric Identities Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 10 Maths Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

#### Page No 617:

We know
${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$
Squaring on both sides, we get

$\therefore \frac{{\mathrm{sin}}^{4}\theta +{\mathrm{cos}}^{4}\theta }{1-2{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }=\frac{1-2{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }{1-2{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }=1$

Ans

#### Page No 617:

$\left(\mathrm{sec}A-\mathrm{cos}A\right)\left(\mathrm{cot}A+\mathrm{tan}A\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{\mathrm{cos}A}-\mathrm{cos}A\right)\left(\frac{\mathrm{cos}A}{\mathrm{sin}A}+\frac{\mathrm{sin}A}{\mathrm{cos}A}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1-{\mathrm{cos}}^{2}A}{\mathrm{cos}A}\right)\left(\frac{{\mathrm{cos}}^{2}A+{\mathrm{sin}}^{2}A}{\mathrm{sin}A\mathrm{cos}A}\right)$

Ans

#### Page No 617:

$\frac{\mathrm{cosec}A-\mathrm{sin}A}{\mathrm{cosec}A+\mathrm{sin}A}\phantom{\rule{0ex}{0ex}}=\frac{\frac{1}{\mathrm{sin}A}-\mathrm{sin}A}{\frac{1}{\mathrm{sin}A}+\mathrm{sin}A}\phantom{\rule{0ex}{0ex}}=\frac{\frac{1-{\mathrm{sin}}^{2}A}{\mathrm{sin}A}}{\frac{1+{\mathrm{sin}}^{2}A}{\mathrm{sin}A}}\phantom{\rule{0ex}{0ex}}=\frac{1-{\mathrm{sin}}^{2}A}{1+{\mathrm{sin}}^{2}A}$
$=\frac{\frac{1-{\mathrm{sin}}^{2}A}{{\mathrm{cos}}^{2}A}}{\frac{1+{\mathrm{sin}}^{2}A}{{\mathrm{cos}}^{2}A}}$           (Dividing numerator and denominator by cos2A)

#### Page No 617:

${\mathrm{sin}}^{2}\theta \mathrm{tan}\theta +{\mathrm{cos}}^{2}\theta \mathrm{cot}\theta +2\mathrm{sin}\theta \mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{sin}}^{3}\theta }{\mathrm{cos}\theta }+\frac{{\mathrm{cos}}^{3}\theta }{\mathrm{sin}\theta }+2\mathrm{sin}\theta \mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{sin}}^{4}\theta +{\mathrm{cos}}^{4}\theta +2{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }{\mathrm{sin}\theta \mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\frac{{\left({\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \right)}^{2}}{\mathrm{sin}\theta \mathrm{cos}\theta }$

$=\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }+\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}=\mathrm{tan}\theta +\mathrm{cot}\theta$

#### Page No 617:

$=\frac{2\mathrm{sin}\theta \left(1+\mathrm{sin}\theta \right)}{\left(1-\mathrm{sin}\theta \right)\left(1+\mathrm{sin}\theta \right)}\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{sin}\theta }{1-\mathrm{sin}\theta }$

#### Page No 617:

$\left(1+\mathrm{cot}A+\mathrm{tan}A\right)\left(\mathrm{sin}A-\mathrm{cos}A\right)\phantom{\rule{0ex}{0ex}}=\mathrm{sin}A+\mathrm{cot}A\mathrm{sin}A+\mathrm{tan}A\mathrm{sin}A-\mathrm{cos}A-\mathrm{cot}A\mathrm{cos}A-\mathrm{tan}A\mathrm{cos}A\phantom{\rule{0ex}{0ex}}=\mathrm{sin}A+\frac{\mathrm{cos}A}{\mathrm{sin}A}×\mathrm{sin}A+\mathrm{tan}A\mathrm{sin}A-\mathrm{cos}A-\mathrm{cot}A\mathrm{cos}A-\frac{\mathrm{sin}A}{\mathrm{cos}A}×\mathrm{cos}A$
$=\mathrm{sin}A+\mathrm{cos}A+\mathrm{tan}A\mathrm{sin}A-\mathrm{cos}A-\mathrm{cot}A\mathrm{cos}A-\mathrm{sin}A\phantom{\rule{0ex}{0ex}}=\mathrm{sin}A\mathrm{tan}A-\mathrm{cot}A\mathrm{cos}A$

Ans

Ans

ANS

Hence, LHS= RHS

#### Page No 618:

Hence, L.H.S. = R.H.S.

ANS

ANS

Ans

Ans

Hence, LHS = RHS

Hence, LHS = RHS

Also,

We know

#### Page No 629:

Given:

x = secA + sinA       .....(1)

y = secA – sinA       .....(2)

Adding (1) and (2), we get

$x+y=\mathrm{sec}A+\mathrm{sin}A+\mathrm{sec}A-\mathrm{sin}A\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sec}A=x+y\phantom{\rule{0ex}{0ex}}⇒\mathrm{sec}A=\frac{x+y}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{sec}A}=\frac{2}{x+y}$

Subtracting (2) from (1), we get

We know

#### Page No 629:

$\mathrm{LHS}=\sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{m}}{\sqrt{n}}+\frac{\sqrt{n}}{\sqrt{m}}\phantom{\rule{0ex}{0ex}}=\frac{m+n}{\sqrt{mn}}\phantom{\rule{0ex}{0ex}}=\frac{\left(\mathrm{cos}\theta -\mathrm{sin}\theta \right)+\left(\mathrm{cos}\theta +\mathrm{sin}\theta \right)}{\sqrt{\left(\mathrm{cos}\theta -\mathrm{sin}\theta \right)\left(\mathrm{cos}\theta +\mathrm{sin}\theta \right)}}$
$=\frac{2\mathrm{cos}\theta }{\sqrt{{\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta }}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{2\mathrm{cos}\theta }{\mathrm{cos}\theta }\right)}{\left(\frac{\sqrt{{\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta }}{\mathrm{cos}\theta }\right)}\phantom{\rule{0ex}{0ex}}=\frac{2}{\sqrt{\frac{{\mathrm{cos}}^{2}\theta }{{\mathrm{cos}}^{2}\theta }-\frac{{\mathrm{sin}}^{2}\theta }{{\mathrm{cos}}^{2}\theta }}}\phantom{\rule{0ex}{0ex}}=\frac{2}{\sqrt{1-{\mathrm{tan}}^{2}\theta }}\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}$

#### Page No 629:

$\mathrm{cos}\theta -\mathrm{sin}\theta =\sqrt{2}\mathrm{sin}\theta$
Squaring on both sides, we get

#### Page No 629:

$\mathrm{sec}\theta -\mathrm{tan}\theta =\sqrt{2}\mathrm{tan}\theta$
Squaring on both sides, we get

#### Page No 629:

Given:
We know

Adding (1) and (2), we get
$\mathrm{sec}\theta +\mathrm{tan}\theta +\mathrm{sec}\theta -\mathrm{tan}\theta =p+\frac{1}{p}\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sec}\theta =\frac{{p}^{2}+1}{p}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sec}\theta =\frac{{p}^{2}+1}{2p}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{sec}\theta }=\frac{2p}{{p}^{2}+1}$

Subtracting (2) from (1), we get

Now,

#### Page No 629:

Given:
We know

Adding (1) and (2), we get
$\mathrm{cosec}A+\mathrm{cot}A+\mathrm{cosec}A-\mathrm{cot}A=m+\frac{1}{m}\phantom{\rule{0ex}{0ex}}⇒2\mathrm{cosec}A=\frac{{m}^{2}+1}{m}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cosec}A=\frac{{m}^{2}+1}{2m}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{cosec}A}=\frac{2m}{{m}^{2}+1}$

Subtracting (2) from (1), we get

Now,

#### Page No 629:

Given:
We know

Adding (1) and (2), we get

Subtracting (1) from (2), we get

Dividing (3) by (4), we get
$\frac{\frac{1+{x}^{2}}{2x}}{\frac{1-{x}^{2}}{2x}}=\frac{\mathrm{sec}A}{\mathrm{tan}A}\phantom{\rule{0ex}{0ex}}⇒\frac{1+{x}^{2}}{1-{x}^{2}}=\frac{\frac{1}{\mathrm{cos}A}}{\frac{\mathrm{sin}A}{\mathrm{cos}A}}\phantom{\rule{0ex}{0ex}}⇒\frac{1+{x}^{2}}{1-{x}^{2}}=\frac{1}{\mathrm{sin}A}\phantom{\rule{0ex}{0ex}}⇒\frac{1+{x}^{2}}{1-{x}^{2}}=\mathrm{cosec}A$

(b) 1

#### Page No 630:

Hence, the correct answer is option (c).

(b) cos A

#### Page No 631:

$\left(1+\mathrm{tan}\theta +\mathrm{sec}\theta \right)\left(1+\mathrm{cot}\theta -\mathrm{cosec}\theta \right)\phantom{\rule{0ex}{0ex}}=1+\mathrm{cot}\theta -\mathrm{cosec}\theta +\mathrm{tan}\theta +\mathrm{tan}\theta \mathrm{cot}\theta -\mathrm{tan}\theta \mathrm{cosec}\theta +\mathrm{sec}\theta +\mathrm{sec}\theta \mathrm{cot}\theta -\mathrm{sec}\theta \mathrm{cosec}\theta \phantom{\rule{0ex}{0ex}}=1+\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }-\frac{1}{\mathrm{sin}\theta }+\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }+\mathrm{tan}\theta ×\frac{1}{\mathrm{tan}\theta }-\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }×\frac{1}{\mathrm{sin}\theta }+\frac{1}{\mathrm{cos}\theta }+\frac{1}{\mathrm{cos}\theta }×\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }-\frac{1}{\mathrm{cos}\theta \mathrm{sin}\theta }$
$=1+\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }-\frac{1}{\mathrm{sin}\theta }+\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }+1-\frac{1}{\mathrm{cos}\theta }+\frac{1}{\mathrm{cos}\theta }+\frac{1}{\mathrm{sin}\theta }-\frac{1}{\mathrm{cos}\theta \mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}=2+\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }+\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }-\frac{1}{\mathrm{cos}\theta \mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}=2+\frac{{\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta }{\mathrm{cos}\theta \mathrm{sin}\theta }-\frac{1}{\mathrm{cos}\theta \mathrm{sin}\theta }$

Hence, the correct answer is option (d).

#### Page No 631:

$\mathrm{sin}\theta -\mathrm{cos}\theta =0\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta =\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }=1\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}\theta =\mathrm{tan}45°$
$⇒\theta =45°\phantom{\rule{0ex}{0ex}}\therefore {\mathrm{sin}}^{4}\theta +{\mathrm{cos}}^{4}\theta \phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{4}45°+{\mathrm{cos}}^{4}45°\phantom{\rule{0ex}{0ex}}={\left(\frac{1}{\sqrt{2}}\right)}^{4}+{\left(\frac{1}{\sqrt{2}}\right)}^{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}+\frac{1}{4}$
$=\frac{2}{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

Hence, the correct answer is option (b).

#### Page No 631:

$\mathrm{cos}9\alpha =\mathrm{sin}\alpha \phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}9\alpha =\mathrm{cos}\left(90°-\alpha \right)\phantom{\rule{0ex}{0ex}}⇒9\alpha =90°-\alpha \phantom{\rule{0ex}{0ex}}⇒10\alpha =90°$
$⇒5\alpha =\frac{90°}{2}=45°\phantom{\rule{0ex}{0ex}}\therefore \mathrm{tan}5\alpha =\mathrm{tan}45°=1$

Thus, the value of tan5α is 1.

Hence, the correct answer is option (c).

#### Page No 631:

Hence, the correct answer is option (d).

#### Page No 631:

(b) (sec A − tan A)

#### Page No 631:

$\sqrt{\frac{1+\mathrm{cos}\left(A\right)}{1-\mathrm{cos}\left(A\right)}}=\sqrt{\frac{1+\mathrm{cos}\left(A\right)}{1-\mathrm{cos}\left(A\right)}×\frac{1+\mathrm{cos}\left(A\right)}{1+\mathrm{cos}\left(A\right)}}=\frac{1+\mathrm{cos}\left(A\right)}{\sqrt{1-{\mathrm{cos}}^{2}\left(A\right)}}\phantom{\rule{0ex}{0ex}}=\frac{1+\mathrm{cos}\left(A\right)}{\sqrt{{\mathrm{sin}}^{2}\left(A\right)}}=\frac{1+\mathrm{cos}\left(A\right)}{\mathrm{sin}\left(A\right)}=\frac{1}{\mathrm{sin}\left(A\right)}+\frac{\mathrm{cos}\left(A\right)}{\mathrm{sin}\left(A\right)}\phantom{\rule{0ex}{0ex}}=\mathrm{cosec}\left(A\right)+\mathrm{cot}\left(A\right).$
Hence, the correct answer is option B.

#### Page No 631:

(b) $\frac{1-\mathrm{cos\theta }}{1+\mathrm{cos\theta }}$

${\left(\text{cosec}\theta -\text{cot}\theta \right)}^{2}\phantom{\rule{0ex}{0ex}}\text{=}{\left(\frac{1}{\text{sin}\theta }-\frac{\text{cos}\theta }{\text{sin}\theta }\right)}^{2}\phantom{\rule{0ex}{0ex}}\text{=}{\left(\frac{1-\text{cos}\theta }{\text{sin}\theta }\right)}^{2}\phantom{\rule{0ex}{0ex}}\text{=}\frac{{\left(1-\text{cos}\theta \right)}^{2}}{{\text{sin}}^{2}\theta }\phantom{\rule{0ex}{0ex}}\text{=}\frac{{\left(1-\text{cos}\theta \right)}^{2}}{\left(1-{\text{cos}}^{2}\theta \right)}\phantom{\rule{0ex}{0ex}}\text{=}\frac{{\left(1-\text{cos}\theta \right)}^{2}}{\left(1+\text{cos}\theta \right)\left(1-\text{cos}\theta \right)}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left(1-\text{cos}\theta \right)}{\left(1+\text{cos}\theta \right)}\phantom{\rule{0ex}{0ex}}$

#### Page No 631:

(d) 23
We have (tan θ +cot θ) = 5
Squaring both sides, we get:
(tan θ +cot θ)2 = 52
⇒ tan2 θ + cot2 θ + 2 tan θ cot θ = 25
⇒ tan2 θ + cot2 θ + 2 = 25       [∵ tan θ = ]
⇒ tan2 θ + cot2 θ = 25 − 2 = 23

#### Page No 631:

(c) $\frac{b+a}{b-a}$

#### Page No 631:

(d) $\frac{3}{4}$

=

#### Page No 632:

Now,

Hence, the correct answer is option (a).

#### Page No 632:

Hence, the correct answer is option (c).

#### Page No 632:

In ∆ABC,

∠A + ∠B + ∠C = 180º                (Angle sum property of triangle)

⇒ ∠A + ∠B + 90º = 180º            (∠C = 90º)

⇒ ∠A + ∠B = 180º − 90º = 90º

∴ cos(A + B) = cos90º = 0

Thus, the value of cos(A + B) is 0.

Hence, the correct answer is option (a).

(a) 1

#### Page No 632:

(d) 9

We have .

Dividing the numerator and denominator of the given expression by sin θ, we get:

=

= = 9              [∵ 3 cot θ = 4]

#### Page No 632:

(a) $\frac{1}{2}$
Given: 2x = sec A and $\frac{2}{x}$ = tan A
Also, we can deduce that x = and .
So, substituting the values of x and $\frac{1}{x}$ in the given expression, we get:
2$\left({x}^{2}-\frac{1}{{x}^{2}}\right)$ = 2
= 2
=
= $\frac{1}{2}$                  [By using the identity: ]

#### Page No 632:

(c) $\frac{1}{3}$
Given: 3x = cosec θ and $\frac{3}{x}$ = cot θ
Also, we can deduce that x = and .
So, substituting the values of x and $\frac{1}{x}$ in the given expression, we get:
3$\left({x}^{2}-\frac{1}{{x}^{2}}\right)$ = 3
= 3
=
= $\frac{1}{3}$       [By using the identity: ]

#### Page No 632:

(d) 2

Let us first draw a right $∆$ABC right angled at B and $\angle \mathrm{A}=\theta$.
Given: tan θ = $\sqrt{3}$
But tan θ = $\frac{BC}{AB}$
So, $\frac{BC}{AB}$ = $\frac{\sqrt{3}}{1}$
Thus, BC = $\sqrt{3}$k and AB = k

Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2 = ($\sqrt{3}$ k)2 + (k)2
⇒ AC2= 4k2
⇒ AC = 2k
∴ sec θ =

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