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#### Question 1:

If sin $\mathrm{\theta }=\frac{\sqrt{3}}{2}$, find the value of all T-ratios of θ.

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that sin $\theta$ = $\frac{\mathrm{perpendicular}}{\mathrm{hypotenuse}}$= $\frac{AB}{AC}$ = .

So, if AB = $\sqrt{3}k$, then AC = 2k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ BC2 = AC2 $-$ AB2 = (2k)2 $-$ ($\sqrt{3}k$)2
⇒ BC2 = 4k2 $-$ 3k2 = k2
⇒ BC = k
Now, finding the other T-ratios using their definitions, we get:
cos $\theta$  = $\frac{BC}{AC}$ =
tan $\theta$  =

∴ cot $\theta$  = , cosec $\theta$ = and sec $\theta$  =

#### Question 2:

If cos  find the values of all T-ratios of θ.

Let us first draw a right $∆$ABC, right angled at B and .
Now, we know that cos $\theta$ = = = .

So, if BC = 7k, then AC = 25k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 $-$ BC2 = (25k)2 $-$ (7k)2.
⇒ AB2 = 625k2 $-$ 49k2 = 576k2
⇒ AB = 24k
Now, finding the other trigonometric ratios using their definitions, we get:
sin $\theta$ = =
tan $\theta$ =
∴ cot $\theta$ = , cosec $\theta$ = and sec $\theta$  =

#### Question 3:

If tan $\mathrm{\theta }=\frac{15}{8}$ find the values of all T-ratios of θ.

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that tan $\theta$ = $\frac{\mathrm{Perpendicular}}{\mathrm{Base}}$ = $\frac{AB}{BC}$ = $\frac{15}{8}$.

So, if BC = 8k, then AB = 15k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (15k)2 + (8k)2
⇒ AC2 = 225k2 + 64k2 = 289k2
⇒ AC = 17k

Now, finding the other T-ratios using their definitions, we get:
sin $\theta$  = $\frac{AB}{AC}$ =
cos $\theta$  =

∴ cot $\theta$  = , cosec $\theta$ = and sec $\theta$  =

#### Question 4:

If cot θ = 2, find the value of all T-ratios of θ.

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that cot $\theta$$\frac{\mathrm{base}}{\mathrm{Perpendicular}}$ = $\frac{BC}{AB}$ = 2.

So, if BC = 2k, then AB = k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (2k)2 + (k)2
⇒ AC2 = 4k2 + k2 = 5k2
⇒ AC = $\sqrt{5}$k
Now, finding the other T-ratios using their definitions, we get:
sin $\theta$  = $\frac{AB}{AC}$ =
cos $\theta$  =

∴ tan $\theta$  = , cosec $\theta$ = and sec $\theta$  =

#### Question 5:

If cosec θ = $\sqrt{10}$, the find the values of all T-ratios of θ.

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that cosec $\theta$ = $\frac{\mathrm{Hypotenuse}}{\mathrm{Perpendicular}}$ = $\frac{AC}{AB}$= $\frac{\sqrt{10}}{1}$.

So, if AC = ($\sqrt{10}$)k, then AB = k, where k is a positive number.
Now, by using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ BC2 = AC2 $-$ AB2 = 10k2 $-$ k2
⇒ BC2 = 9k2
⇒ BC = 3k
Now, finding the other T-ratios using their definitions, we get:
tan $\theta$  = $\frac{AB}{BC}$ =

cos $\theta$  =

∴ , cot $\theta$  = and sec $\theta$  =

#### Question 6:

If cot θ = $\frac{15}{8}$, show then evaluate

Let us consider a right $△$ABC, right angled at B.
Now, we know that cot $\theta$ =$\frac{\mathrm{Base}}{\mathrm{Perpendicular}}$ =$\frac{BC}{AB}$ = $\frac{15}{8}$.

Using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (8k)2 + (15k)2
⇒ AC2 = 64k2 + 225k2 = 289k2
⇒ AC = 17k

Finding out the value of sin $\theta$ and cos $\theta$ using their definitions, we have:
sin $\theta$ =
cos $\theta$ =

Substituting these values in the given expression, we get:

#### Question 7:

If tan θ = $\frac{4}{3}$, show that (sin θ + cos θ) = $\frac{7}{5}$.

Let us consider a right $△$ABC, right angled at B and $\angle C=\theta$.
Now, we know that  tan $\theta$$\frac{AB}{BC}$$\frac{4}{3}$.

So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (4k)2 + (3k)2
⇒ AC2 = 16k2 + 9k2 = 25k2
⇒ AC = 5k
Finding out the values of sin $\theta$ and cos $\theta$ using their definitions, we have:
sin $\theta$ =
cos $\theta$ =
Substituting these values in the given expression, we get:
(sin $\theta$ + cos $\theta$) =
i.e., LHS = RHS

Hence proved.

#### Question 8:

If tan θ = $\frac{1}{\sqrt{7}}$ then prove that $\left(\frac{{\mathrm{cosec}}^{2}\mathrm{\theta }+{\mathrm{sec}}^{2}\mathrm{\theta }}{{\mathrm{cosec}}^{2}\mathrm{\theta }-{\mathrm{sec}}^{2}\mathrm{\theta }}\right)=\frac{4}{3}.$

Let us consider a right $△$ABC, right angled at B and $\angle C=\theta$.
Now it is given that tan $\theta$$\frac{AB}{BC}$$\frac{1}{\sqrt{7}}$.

So, if AB = k, then BC = $\sqrt{7}$k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AC2 = (k)2 + ($\sqrt{7}$k)2
⇒ AC2 = k2 + 7k2
⇒ AC = 2$\sqrt{2}$k
Now, finding out the values of the other trigonometric ratios, we have:
sin $\theta$  =
cos $\theta$  =
∴ cosec $\theta$  = and sec $\theta$   =
Substituting the values of cosec $\theta$  and sec $\theta$  in the given expression, we get:

i.e., LHS = RHS

Hence proved.

#### Question 9:

If cosec θ = 2, show that $\left(\mathrm{cot\theta }+\frac{\mathrm{sin\theta }}{1+\mathrm{cos\theta }}\right)=2.$

Let us consider a right $△$ABC, right angled at B and $\angle C=\theta$.
Now, it is given that cosec $\theta$ = 2.
Also, sin $\theta$  =

So, if AB = k, then AC = 2k, where k is a positive number.
Using Pythagoras theorem, we have:
⇒ AC2 = AB2  + BC2
⇒ BC2 = AC2 $-$ AB2
⇒ BC2 = (2k)2 $-$ (k)2
⇒ BC2 = 3k2
⇒ BC = $\sqrt{3}$k
Finding out the other T-ratios using their definitions, we get:
cos $\theta$ =
tan $\theta$ =
cot $\theta$ =
Substituting these values in the given expression, we get:

$=\sqrt{3}+\frac{1}{2+\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}\left(2+\sqrt{3}\right)+1}{2+\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{3}+3+1}{2+\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2\left(2+\sqrt{3}\right)}{2+\sqrt{3}}=2$
i.e., LHS = RHS

Hence proved.

#### Question 10:

If sec θ = $\frac{5}{4}$, verify that $\frac{\mathrm{tan\theta }}{\left(1+{\mathrm{tan}}^{2}\mathrm{\theta }\right)}=\frac{\mathrm{sin\theta }}{sec\mathrm{\theta }}$.

Let us consider a right $△$ABC, right angled at B and $\angle C=\theta$.
Now, it is given that sec $\theta$ = $\frac{5}{4}$.
Also, cos $\theta$  =

So, if BC = 4k, then AC = 5k, where k is a positive number.
Using Pythagoras theorem, we have:
⇒ AC2 = AB2  + BC2
⇒ AB2 = AC2 $-$ BC2
⇒ AB2 = (5k)2 $-$ (4k)2 = 25k2 $-$ 16k2
⇒ AB2 = 9k2
⇒ AB = 3k

Finding out the other T-ratios using their definitions, we get:
sin $\theta$ =

tan $\theta$ =

Substituting the values in LHS and RHS, we get:

i.e., LHS = RHS

Hence proved.

#### Question 11:

If cos θ = 0.6, show that (5 sin θ − 3 tan θ) = 0.

Let us consider a right $△$ABC right angled at B.
Now, we know that cos $\theta$ = 0.6 = $\frac{BC}{AC}$$\frac{3}{5}$

So, if BC = 3k, then AC = 5k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2  + BC2
⇒ AB2 = AC2 $-$ BC2
⇒ AB2 = (5k)2 $-$ (3k)2 = 25k2 $-$ 9k2
⇒ AB2 = 16k2
⇒ AB = 4k

Finding out the other T-ratios using their definitions, we get:
sin $\theta$ =

tan $\theta$ =

Substituting the values in the given expression, we get:
5 sin $\theta$ $-$ 3 tan $\theta$

i.e., LHS  = RHS

Hence proved.

#### Question 12:

If 3 tan θ = 4, show that $\frac{4\mathrm{cos\theta }-\mathrm{sin\theta }}{2\mathrm{cos\theta }+\mathrm{sin\theta }}=\frac{4}{5}.$

Let us consider a right $△$ABC right angled at B and $\angle C=\theta$.
We know that tan $\theta$ = $\frac{AB}{BC}$$\frac{4}{3}$

So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2  + BC2
⇒ AC2 = 16k2 + 9k2
⇒ AC2 = 25k2
⇒ AC = 5k

Now, we have:

sin $\theta$ =

cos $\theta$ =

Substituting these values in the given expression, we get:

i.e., LHS = RHS

Hence proved.

#### Question 13:

If tan θ = $\frac{a}{b}$, show that

It is given that tan .

LHS =
Dividing the numerator and denominator by cos $\theta$, we get:

(∵ tan )
Now, substituting the value of tan $\theta$ in the above expression, we get:

i.e., LHS = RHS

Hence proved.

#### Question 14:

If 3 cot θ = 2, show that $\left(\frac{4\mathrm{sin\theta }-3\mathrm{cos\theta }}{2\mathrm{sin\theta }+6\mathrm{cos\theta }}\right)=\frac{1}{3}.$

It is given that cot .

LHS  =
Dividing the above expression by sin $\theta$, we get:
[∵ cot ]
Now, substituting the values of cot $\theta$ in the above expression, we get:

i.e., LHS = RHS

Hence proved.

#### Question 15:

If sec θ = $\frac{17}{8}$ then prove that $\frac{3-4{\mathrm{sin}}^{2}\mathrm{\theta }}{4{\mathrm{cos}}^{2}\mathrm{\theta }-3}=\frac{3-{\mathrm{tan}}^{2}\mathrm{\theta }}{1-3{\mathrm{tan}}^{2}\mathrm{\theta }}$.

It is given that sec $\theta$ = $\frac{17}{8}$.

Let us consider a right $△$ABC right angled at B and $\angle C=\theta$.
We know that cos $\theta$ =

So, if BC = 8k, then AC = 17k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 $-$ BC2 = (17k)2 $-$ (8k)2
⇒ AB2 = 289k2 $-$ 64k2 = 225k2
⇒ AB = 15k.

Now, tan $\theta$  = and sin $\theta$ =

The given expression is .

Substituting the values in the above expression, we get:

∴ LHS = RHS
Hence proved.

#### Question 16:

If tan θ = $\frac{20}{21}$, show that$\frac{\left(1-\mathrm{sin\theta }+\mathrm{cos\theta }\right)}{\left(1+\mathrm{sin\theta }+\mathrm{cos\theta }\right)}=\frac{3}{7}.$

Let us consider a right $△$ABC right angled at B and $\angle C=\theta$.
Now, we know that tan $\theta$$\frac{AB}{BC}$ = $\frac{20}{21}$

So, if AB = 20k, then BC = 21k, where k is a positive number.
Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2= (20k)2 + (21k)2
⇒ AC2 = 841k2
⇒  AC = 29k
Now, sin $\theta$ = and cos $\theta$ =

Substituting these values in the given expression, we get:

∴ LHS = RHS

Hence proved.

#### Question 17:

In a ∆ABC, right-angled at B, it is given that AB = 12 cm and BC = 5 cm.

Find the value of
(i) cos A
(ii) cosec A
(iii) cos C
(iv) coses C.

Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2 = 122  + 52 = 144 + 25
⇒ AC2 = 169
⇒ AC = 13 cm
Now, for T-Ratios of ∠A, base = AB and perpendicular = BC
(i) cos A =
(ii) cosec A =

Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) cos C =
(iv) cosec C =

#### Question 18:

In a ∆ABC, it is given that ∠B = 90°, AB = 24 cm and BC = 7 cm.
Figure

Find the value of
(i) sin A
(ii) cos A
(iii) sin C
(iv) cos C

Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2= (24)2 + (7)2
⇒ AC2 = 576 + 49 = 625
⇒ AC = 25 cm
Now, for T-Ratios of ∠A, base = AB and perpendicular = BC
(i)  sin A =

(ii) cos A =

Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) sin C =

(iv) cos C =

#### Question 19:

Given a ∆ABC, in which ∠C = 90°, ∠ABC = θ°, BC = 21 units, AB = 29 units.
Show that (cos2θ − sin2θ) = $\frac{41}{841}.$

Using Pythagoras theorem, we get:
AB2  =  AC2  + BC2
⇒ AC2 = AB2 $-$ BC2
⇒ AC2 = (29)2 $-$ (21)2
⇒ AC2 = 841$-$ 441
⇒ AC2 = 400
⇒ AC = $\sqrt{400}$ = 20 units

Now, sin and cos $\theta$ =

cos2 $\theta$ $-$ sin2 $\theta$ =

Hence Proved.

#### Question 1:

If cosθ = $\frac{4}{5}$, then tanθ = ?
(a) $\frac{3}{4}$
(b) $\frac{4}{3}$
(c) $\frac{3}{5}$
(d) $\frac{5}{3}$

(a) $\frac{3}{4}$
Since cosθ = $\frac{4}{5}$ but cosθ = $\frac{AB}{AC}$
So, $\frac{AB}{AC}$ = $\frac{4}{5}$
Thus, AB = 4k and AC = 5k

Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ BC2 = AC2 $-$ AB2
⇒ BC2  = (5k)2 $-$ (4k)2
⇒ BC2 = 9k2
⇒ BC = 3k
∴ tanθ = $\frac{BC}{AB}$ = $\frac{3}{4}$

#### Question 2:

If sinθ = $\frac{1}{2}$, then cotθ = ?
(a) $\frac{1}{\sqrt{3}}$
(b) $\sqrt{3}$
(c) $\frac{\sqrt{3}}{2}$
(d) 1

(b) $\sqrt{3}$

Given: sinθ = $\frac{1}{2}$, but sinθ = $\frac{\mathrm{BC}}{\mathrm{AC}}$
So, $\frac{\mathrm{BC}}{\mathrm{AC}}$ = $\frac{1}{2}$
Thus, BC = k and AC = 2k

Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
AB2 = AC2 $-$ BC2
AB2 = (2k)2 $-$ (k)2
AB2 = 3k2
AB = $\sqrt{3}$k
So, tanθ = $\frac{\mathrm{BC}}{\mathrm{AB}}$ = $\frac{k}{\sqrt{3}k}$ = $\frac{1}{\sqrt{3}}$
∴ cotθ = $\frac{1}{\mathrm{tan}\theta }$ = $\sqrt{3}$

#### Question 3:

If tan θ = $\frac{3}{4}$, then cosec θ = ?
(a) $\frac{4}{3}$
(b) $\frac{4}{5}$
(c) $\frac{5}{4}$
(d) $\frac{5}{3}$

(d) $\frac{5}{3}$

Given: tan $\theta$ = $\frac{3}{4}$, but tan $\theta$ = $\frac{\mathrm{Perpendicular}}{\mathrm{Base}}$
So, let us first draw a right $∆$ABC right angled at B and $\angle A=\theta$.
So, $\frac{BC}{AB}$ = $\frac{3}{4}$
Thus, BC = 3k and AB = 4k

Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ AC2 = (4k)2 + (3k)2
⇒ AC2 = 25k2
⇒ AC = 5k
∴ cosec $\theta$ = $\frac{AC}{BC}$ = $\frac{5k}{3k}$ = $\frac{5}{3}$

#### Question 4:

If cosec θ = $\sqrt{10}$, then sec θ = ?
(a) $\frac{3}{\sqrt{10}}$
(b) $\frac{\sqrt{10}}{3}$
(c) $\frac{1}{\sqrt{10}}$
(d) $\frac{2}{\sqrt{10}}$

(b) $\frac{\sqrt{10}}{3}$
Let us first draw a right $∆$ABC right angled at B and $\angle A=\theta$.
Given: cosec $\theta$$\sqrt{10}$, but sin $\theta$ = $\frac{1}{\sqrt{10}}$
Also, sin $\theta$ = $\frac{\mathrm{Perpendicular}}{\mathrm{Hypotenuse}}$ = $\frac{BC}{AC}$
So, $\frac{BC}{AC}$ = $\frac{1}{\sqrt{10}}$
Thus, BC = k and AC = $\sqrt{10}$ k

Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
AB2 = AC2 $-$ BC2
AB2 = ($\sqrt{10}$ k)2 $-$ (k)2
AB2 = 9k2
AB = 3k
∴ sec $\theta$ = $\frac{AC}{AB}$ = $\frac{\sqrt{10}k}{3k}=\frac{\sqrt{10}}{3}$

#### Question 5:

If tan θ = $\frac{8}{15}$, then cosec θ = ?
(a) $\frac{17}{8}$
(b) $\frac{8}{17}$
(c) $\frac{17}{15}$
(d) $\frac{15}{17}$

(a) $\frac{17}{8}$

Let us first draw a right $∆$ABC right angled at B and $\angle A=\theta$.
Given: tan $\theta$ = $\frac{8}{15}$, but tan $\theta$ = $\frac{BC}{AB}$
So, $\frac{BC}{AB}$ = $\frac{8}{15}$
Thus, BC = 8k and AB = 15k

Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ AC2 = (15k)2 + (8k)2
⇒ AC2 = 289k2
⇒ AC = 17k
∴ cosec $\theta$ = $\frac{AC}{BC}=\frac{17k}{8k}=\frac{17}{8}$

#### Question 6:

If sin θ $\frac{a}{b}$, then cos θ = ?
(a) $\frac{b}{\sqrt{{b}^{2}-{a}^{2}}}$
(b) $\frac{\sqrt{{b}^{2}-{a}^{2}}}{b}$
(c) $\frac{a}{\sqrt{{b}^{2}-{a}^{2}}}$
(d) $\frac{b}{a}$

(b) $\frac{\sqrt{{b}^{2}-{a}^{2}}}{b}$
Let us first draw a right $∆$ABC right angled at B and $\angle A=\theta$.
Given: sin θ = $\frac{a}{b}$, but sin θ = $\frac{BC}{AC}$
So, $\frac{BC}{AC}$ = $\frac{a}{b}$
Thus, BC = ak and AC = bk

Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 $-$ BC2
⇒ AB2 = (bk)2 $-$ (ak)2
⇒ AB2 = $\left({b}^{2}-{a}^{2}\right)$k2
⇒ AB = ($\sqrt{{b}^{2}-{a}^{2}}$)k
∴ cos θ  = $\frac{\mathrm{AB}}{\mathrm{AC}}$ = $\frac{\sqrt{{b}^{2}-{a}^{2}}k}{bk}$ = $\frac{\sqrt{{b}^{2}-{a}^{2}}}{b}$

#### Question 7:

If tan θ = $\sqrt{3}$, then sec θ = ?
(a) $\frac{2}{\sqrt{3}}$
(b) $\frac{\sqrt{3}}{2}$
(c) $\frac{1}{2}$
(d) 2

(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 =

#### Question 8:

If sec θ = $\frac{25}{7}$, then sin θ = ?
(a) $\frac{7}{24}$
(b) $\frac{24}{7}$
(c) $\frac{24}{25}$
(d) None of these

(c) $\frac{24}{25}$
Let us first draw a right $∆$ABC right angled at B and $\angle A=\theta$.
Given: sec θ = $\frac{25}{7}$
But cos θ$\frac{AB}{AC}$ = $\frac{7}{25}$
Thus, AC = 25k and AB = 7k

Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
BC2 = AC2 $-$AB2
BC2 = (25k)2 $-$ (7k)2
BC2= 576k2
BC = 24k

∴ sin θ =

#### Question 9:

If sin θ = $\frac{\sqrt{3}}{2}$, then (cosecθ + cotθ) = ?
(a) $2\sqrt{3}$
(b) $\frac{2\sqrt{3}}{3}$
(c) $\left(2+\sqrt{3}\right)$
(d) $\sqrt{3}$

(d) $\sqrt{3}$
Let us first draw a right $∆$ABC right angled at B and $\angle A=\theta$.
sin θ = $\frac{\sqrt{3}}{2}$ = $\frac{BC}{AC}$
So, BC = $\sqrt{3}$k and AC = 2k

Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
AB2 = AC2 $-$ BC2
AB2 = (2k)2  $-$ ($\sqrt{3}$k)2
AB2 = k2
AB  = k

So, cot θ = $\frac{AB}{BC}$ =
cosec θ =
∴ (cosec θ + cot θ) = =

#### Question 10:

If tan θ = $\frac{4}{3}$, then (sinθ + cosθ) = ?
(a) $\frac{7}{3}$
(b) $\frac{7}{4}$
(c) $\frac{7}{5}$
(d) $\frac{5}{7}$

(c) $\frac{7}{5}$
Let us first draw a right $∆$ABC right angled at B and $\angle A=\theta$.
tan θ =
So, AB = 3k and BC = 4k

Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
AC2 = (3k)2 + (4k)2
AC2 = 25k2
AC= 5k
Thus, sin θ = $\frac{BC}{AC}$ = $\frac{4}{5}$
and cos θ =
∴ (sin θ + cos θ) = ( $\frac{4}{5}$ + $\frac{3}{5}$) = $\frac{7}{5}$

#### Question 11:

If 3x = cosec θ and $\frac{3}{x}$ = cot θ, than $3\left({x}^{2}-\frac{1}{{x}^{2}}\right)=?$
(a) $\frac{1}{27}$
(b) $\frac{1}{81}$
(c) $\frac{1}{3}$
(d) $\frac{1}{9}$

(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: ]

#### Question 12:

If 2x = sec A and $\frac{2}{x}$ = tan A, then $2\left({x}^{2}-\frac{1}{{x}^{2}}\right)=?$ =?
(a) $\frac{1}{2}$
(b) $\frac{1}{4}$
(c) $\frac{1}{8}$
(d) $\frac{1}{16}$

(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: ]

#### Question 13:

If (tan θ + cot θ) = 5, then (tan2 θ + cot2 θ) = ?

(a) 23
(b) 24
(c) 25
(d) 27

(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

#### Question 14:

If (cos θ + sec θ) = $\frac{5}{2}$, then (cos2 θ + sec2 θ) = ?
(a) $\frac{21}{4}$
(b) $\frac{17}{4}$
(c) $\frac{29}{4}$
(d) $\frac{33}{4}$

(b) $\frac{17}{4}$
We have (cos θ +sec θ) = $\frac{5}{2}$
Squaring both sides, we get:
(cos θ + sec θ)2 = ($\frac{5}{2}$)2
$⇒$cos2 θ + sec2 θ + 2 cos θ sec θ = $\frac{25}{4}$
$⇒$cos2 θ + sec2 θ + 2 = $\frac{25}{4}$      [∵ sec θ = ]
$⇒$cos2 θ + sec2 θ = $\frac{25}{4}$ − 2 = $\frac{17}{4}$

#### Question 15:

If tan θ = $\frac{1}{\sqrt{7}}$, then $\frac{\left(\mathrm{cos}e{c}^{2}\mathrm{\theta }-se{c}^{2}\mathrm{\theta }\right)}{\left(\mathrm{cos}e{c}^{2}\mathrm{\theta }+se{c}^{2}\mathrm{\theta }\right)}=?$ = ?
(a) $\frac{-2}{3}$
(b) $\frac{-3}{4}$
(c) $\frac{2}{3}$
(d) $\frac{3}{4}$

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

=

#### Question 16:

If 7 tan θ = 4, then $\frac{\left(7\mathrm{sin\theta }-3\mathrm{cos\theta }\right)}{\left(7\mathrm{sin\theta }+3\mathrm{cos\theta }\right)}$ = ?
(a) $\frac{1}{7}$
(b) $\frac{5}{7}$
(c) $\frac{3}{7}$
(d) $\frac{5}{14}$

(a) $\frac{1}{7}$

7 tan θ = 4

Now, dividing the numerator and denominator of the given expression by cos θ, we get:

=

=         [∵ 7 tan θ = 4]
$\frac{1}{7}$

#### Question 17:

If 3 cot θ = 4, then $\frac{\left(5\mathrm{sin\theta }+3\mathrm{cos\theta }\right)}{\left(5\mathrm{sin\theta }-3\mathrm{cos\theta }\right)}$ = ?
(a) $\frac{1}{3}$
(b) 3
(c) $\frac{1}{9}$
(d) 9

(d) 9

We have .

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

=

= = 9              [∵ 3 cot θ = 4]

#### Question 18:

If θ = $\frac{a}{b}$, then  = ?
(a) $\frac{\left({a}^{2}+{b}^{2}\right)}{\left({a}^{2}-{b}^{2}\right)}$
(b) $\frac{\left({a}^{2}-{b}^{2}\right)}{\left({a}^{2}+{b}^{2}\right)}$
(c) $\frac{{a}^{2}}{\left({a}^{2}+{b}^{2}\right)}$
(d) $\frac{{b}^{2}}{\left({a}^{2}+{b}^{2}\right)}$

(b)
We have tan θ = $\frac{a}{b}$

Now, dividing the numerator and denominator of the given expression by cos θ, we get:

#### Question 19:

If cot θ = , then $\frac{\left(2+2\mathrm{sin\theta }\right)\left(1-\mathrm{sin\theta }\right)}{\left(1+\mathrm{cos\theta }\right)\left(2-2\mathrm{cos\theta }\right)}$ = ?
(a) $\frac{64}{225}$
(b) $\frac{225}{64}$
(c) $\frac{64}{289}$
(d) $\frac{289}{64}$

(b) $\frac{225}{64}$
We have:

#### Question 20:

In the given ∆ABC, it is given that ∠B = 90°, AB = 24 cm and BC = 7 cm. Then, cos A = ?
Figure

(a) $\frac{7}{24}$
(b) $\frac{7}{25}$
(c) $\frac{24}{25}$
(d) None of thse

(c) $\frac{24}{25}$

Here, AB = 24 cm and BC = 7 cm
Now, using Pythagoras theorem, we get:
AC2 = AB2 + BC2
AC2 = (24)2 + (7)2
AC2 = 625
AC = 25 cm
∴ cos A =

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