NCERT Exemplar Class 10 Maths Solutions Chapter 8 - Introduction To Trigonometry And Its Equations

Question:1

If cos A =4/5 , then the value of tan A is

1.$\frac{3}{5}$ (b) $\frac{3}{4}$ (c) $\frac{4}{3}$ (d) $\frac{5}{3}$

Answer:

Answer. [B]
Solution. It is given that cos A = 4/5
$\text{We know that cos}\theta =\frac{Base}{Hypotenuse}$
$\therefore$ value of base = 4
Hypotenuse = 5

Use Pythagoras theorem in $△$ABC
(Hypotenuse)² = (Base)² + (perpendicular)
${\left(AC\right)}^2={\left(BC\right)}^2+{\left(AB\right)}^2$
${\left(5\right)}^2={\left(4\right)}^2+{\left(AB\right)}^2$
$25-16={\left(AB\right)}^2$
$9={\left(AB\right)}^2$
$\sqrt{9}=AB$
$3=AB$
The value of perpendicular is 3
$\text{Also we know that tan}\theta =\frac{perpendicular}{base}$
$\therefore \tan A=\frac{3}{4}$
Hence, option (B) is correct.

Question:2

If sin A =1/2, then the value of cot A is
a) $\sqrt{3}$ (b) $\frac{1}{\sqrt{3}}$ (c) $\frac{\sqrt{3}}{2}$ (d) 1

Answer:

$$\begin{gathered} sinA=\frac{1}{2}\text{ for angle }{30}^0 \\ cosA=\frac{\sqrt{3}}{2} \\ cotA=\frac{cosA}{sinA}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3} \end{gathered}$$
Hence, option (a) is correct.

Question:3

The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is
(A) – 1 (B) 0 (C) 1 (D)$\frac{3}{2}$

Answer:

Answer. $[B]$
Solution. Given expression is :

$\begin{array}{rl}& [\mathrm{cosec}({75}^{\circ }+\theta )-\sec ({15}^{\circ }-\theta )-\tan ({55}^{\circ }+\theta )+\cot ({35}^{\circ }-\theta )]\\ & [\mathrm{cosec}(90-(15-\theta ))-\sec ({15}^{\circ }-\theta )-\tan (90-(35-\theta ))+\cot ({35}^{\circ }-\theta )]\\ & \{\because \text{ we can write }(75+\theta )=(90-(15-\theta ))\text{ and }(55+\theta )=(90-(35-\theta ))\}\\ & \{\because \mathrm{cosec}(90-q)=\sec q\text{ and tan }(90-q)=\cot q\}\\ & [\sec (15-\theta )-\sec (15-\theta )-\cot (35-\theta )+\cot (35-\theta )]=0\end{array}$
Hence, option (B) is correct

Question:4

Given that sinθ =a/b, then cosθ is equal to
(A) $\frac{b}{\sqrt{b^2-a^2}}$ (B) $\frac{b}{a}$ (C) $\frac{\sqrt{b^2-a^2}}{b}$ (D)$\frac{a}{\sqrt{b^2-a^2}}$

Answer:

Answer. [C]
Solution. It is given that sin$\theta$ = a/b
$cos\theta =\sqrt{1-si{n}^2\theta }=\sqrt{1-\frac{a^2}{b^2}}$

$\therefore \cos \theta =\frac{\sqrt{b^2-a^2}}{b}$
Hence option (C) is correct.

Question:5

If cos (α + β) = 0, then sin (α – β) can be reduced to
(A) cos β (B) cos 2β (C) sin α (D) sin 2α

Answer:

$$\begin{gathered} \text{Given that }\cos (\alpha +\beta )=0 \\ \Rightarrow (\alpha +\beta )={90}^0...........(1) \end{gathered}$$

$\text{wehave to find }\sin (\alpha -\beta )$

$\text{from(1) we can write }\alpha =90-\beta$

$\therefore \sin (\alpha -\beta )=\sin (90-\beta -\beta )$

$=\sin (90-2\beta )=\cos 2\beta$

$\text{since sin(90-x)=cosx})$

Hence, option B is correct.

Question:6

The value of (tan1° tan2° tan3° ... tan89°) is
(A) 0 (B) 1 (C) 2 (D)1/2

Answer:

Answer. [B]
Solution. Given :-tan1° tan2° tan3° ... tan89°
tan1° tan2° tan3° ... tan89°tan87° tan 88° tan89° …(1)
We can also write equation (1) in the form of
[tan (90⁰ – 89⁰). tan (90⁰ – 88⁰). tan (90⁰ – 87⁰) …… tan 87°. tan 88° tan 89°]
[$\because$ we can write tan 1° in the form of tan (90⁰ – 89⁰) similarly, we can write other values]
[cot 89⁰. cot 88⁰. cot 87⁰ …. tan 87⁰. tan 88⁰. tan 89⁰]
$\because$ [tan (90² – $\theta$) = cot$\theta$ ]
Also
$[\frac{1}{\tan {89}^{\circ }}\frac{1}{\tan {88}^{\circ }}\frac{1}{\tan {87}^{\circ }}\cdots \tan {87}^{\circ }.\tan {88}^{\circ }\tan {89}^{\circ }]$
$\because$ Throughout all terms are cancelled by each other and remaining will be tan45
Hence, the value is 1
$\because$ option B is correct.

Question:7

If cos 9α = sinα and 9α < 90° , then the value of tan5α is

(A) $\frac{1}{\sqrt{3}}$

(B) $\sqrt{3}$

(C) $1$

(D) $0$

Answer:

Answer. [C]
Solution. Given :- cos 9$\alpha$ = sin$\alpha$
cos9 $\alpha$ = cos(90 – $\alpha$)
$\because$ (cos (90 – $\alpha$) = sin$\alpha$)
9$\alpha$ = 90 – $\alpha$
9$\alpha$+ $\alpha$ = 90
10 $\alpha$= 90
$\alpha =\frac{90}{10}$
Now tan 5$\alpha$ is
Put $\alpha$ = 9 we get
tan 5$\times$ (9)
tan 45⁰
= 1
{$\because$ from the table of trigonometric ratios of angles we know that tan 45⁰ = 1}
Hence, option C is correct.

Question:8

If ΔABC is right-angled at C, then the value of cos (A+B) is
$(A)0(B)1(C)1/2(D)\frac{\sqrt{3}}{2}$

Answer:

Answer. [A]
Solution. It is given that $\mathrm{\angle }$C = 90°

In $△$ABC
$\mathrm{\angle }$A +$\mathrm{\angle }$B +$\mathrm{\angle }$C = 180 [$\because$ sum of interior angles of triangle is 180°]
$\mathrm{\angle }$A + $\mathrm{\angle }$B + 90o = 180 [$\because$ C = 90° (given)]
$\mathrm{\angle }$A + $\mathrm{\angle }$B = 180⁰ – 90⁰
$\mathrm{\angle }$A + $\mathrm{\angle }$B = 90° …(1)
cos($\mathrm{\angle }$A +$\mathrm{\angle }$B) = cos (90°)
cos(90°) = 0
[$\because$ from the table of trigonometric ratios of angles we know that cos 90° = 0]
Hence option A is correct.

Question:9

If sinA + sin²A = 1, then the value of the expression (cos²A+ cos⁴A) is
(A) 1 (B)1/2 (C) 2 (D) 3

Answer:

Answer. [A]
Solution. It is given that sinA + sin²A = 1 …(*)
sinA = 1 – sin²AsinA = cos²A …(1) ( $\because$ 1 – sin²A = cos²A)
Squaring both sides we get
sin² A = cos⁴A …(2)
Hence cos²A + cos⁴A =
= sinA + sin²A {using (1) and (2)}
= sinA + sin²A = 1 (given)
Hence option (A) is correct.

Question:10

Given that sinα =1/2 and cosβ =1/2 , then the value of (α + β) is
(A) 0° (B) 30° (C) 60° (D) 90°

Answer:

Answer. [D]
Solution.
$$\begin{gathered} \sin \alpha =1/2\Rightarrow \alpha ={30}^0 \\ \cos \beta =1/2\Rightarrow \beta ={60}^0 \\ \therefore \alpha +\beta ={90}^0 \end{gathered}$$
Hence option (D) is correct.

Question:11

The value of the expression
$[\frac{{\sin }^2{22}^{\circ }+{\sin }^2{68}^{\circ }}{{\cos }^2{22}^{\circ }+{\cos }^2{68}^{\circ }}+{\sin }^2{63}^{\circ }+{\cos }^2{63}^{\circ }\sin {27}^{\circ }]is$
(A) 3 (B) 2 (C) 1 (D) 0

Answer:

Answer. [B]
Solution.

$[\frac{{\sin }^2{22}^{\circ }+{\sin }^2{68}^{\circ }}{{\cos }^2{22}^{\circ }+{\cos }^2{68}^{\circ }}+{\sin }^2{63}^{\circ }+{\cos }^2{63}^{\circ }\sin {27}^{\circ }]$
$\frac{{\sin }^2{22}^{\circ }+{\sin }^2{68}^{\circ }}{{\cos }^2{22}^{\circ }+{\cos }^2{68}^{\circ }}+{\sin }^2{63}^{\circ }+{\cos }^2{63}^{\circ }\sin ({90}^{\circ }-{63}^{\circ })$
$[\because \sin ({90}^{\circ }-\theta )=\cos \theta ]$
$\frac{{\sin }^2{22}^{\circ }+{\sin }^2{68}^{\circ }}{{\cos }^2{22}^{\circ }+{\cos }^2{68}^{\circ }}+{\sin }^2{63}^{\circ }+\cos {63}^{\circ }\times \cos {63}^{\circ }$
$=\frac{{\sin }^2{22}^{\circ }+{\sin }^2({90}^{\circ }-{22}^{\circ })}{{\cos }^2({90}^{\circ }-{68}^{\circ })+{\cos }^2{68}^{\circ }}+{\sin }^2{63}^{\circ }+{\cos }^2{63}^{\circ }$
$\left\{\begin{array}{cc}\because {68}^{\circ }=({90}^{\circ }-{22}^{\circ })& \\ {22}^{\circ }=({90}^{\circ }-{68}^{\circ })& \end{array}\right\}$
$=\frac{{\sin }^2{22}^{\circ }+{\cos }^2{22}^{\circ }}{{\sin }^2{68}^{\circ }+{\cos }^2{68}^{\circ }}+{\sin }^2{63}^{\circ }+{\cos }^2{63}^{\circ }$
$\left\{\begin{array}{cc}\because \sin ({90}^{\circ }-\theta )=\cos \theta & \\ \cos ({90}^{\circ }-\theta )=\sin \theta & \end{array}\right\}$
$=\frac{1}{1}+1[\because {\sin }^2\theta +{\cos }^2\theta =1]$
= 1+ 1 =2
Hence option (B) is correct.

Question:12

If $4\tan \theta =3$ then $\left(\frac{4\sin \theta -\cos \theta }{4\sin \theta +\cos \theta }\right)$ is equal to
$(A)\frac{2}{3}$ $(B)\frac{1}{3}$ $(C)\frac{1}{2}$ $(D)\frac{3}{4}$

Answer:
Hence option (C) is correct.
$$\begin{gathered} \left(\frac{4\sin \theta -\cos \theta }{4\sin \theta +\cos \theta }\right)\text{ Divide numerator and denominator by and rewriting the given expression} \\ \Rightarrow (\frac{4\sin \theta -\cos \theta }{4\sin \theta +\cos \theta })=\frac{4\frac{\sin \theta }{cos\theta }-\frac{cos\theta }{cos\theta }}{4\frac{\sin \theta }{cos\theta }+\frac{cos\theta }{cos\theta }} \\ =\frac{4\tan \theta -1}{4\tan \theta +1}=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2}(\text{Given }4\tan \theta =3) \end{gathered}$$

Question:13

If sinθ – cosθ = 0, then the value of (sin⁴θ + cos⁴θ) is
(A) 1 (B)3/4 (C) 1/2 (D) 1/4

Answer:

$\sin \theta -\cos \theta =0$
squaring both sides we get
$(\sin \theta -\cos \theta {)}^2=0$
${\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta =0$
($\therefore$ (a – b)² = a² + b² – 2ab)
${\sin }^2\theta +{\cos }^2\theta =2\sin \theta \cos \theta .......(1)$
$1=2\sin \theta \cos \theta \because ({\sin }^2\theta +{\cos }^2\theta =1)$
$\frac{1}{2}=\sin \theta \cos \theta$
Squaring both sides we get
$\frac{1}{4}={\sin }^2\theta {\cos }^2\theta$ …(2)
Now squaring both side of equation (1) we get
${({\sin }^2\theta +{\cos }^2\theta )}^2={(2\sin \theta \cos \theta )}^2$
${({\sin }^2\theta )}^2+{({\cos }^2\theta )}^2+2{\sin }^2\theta \cdot {\cos }^2\theta =4{\sin }^2\theta {\cos }^2\theta$
$[\because {(a+b)}^2=a^2+b^2+2ab]$
$({\sin }^4\theta )+({\cos }^4\theta )=4{\sin }^2\theta {\cos }^2\theta -2{\sin }^2\theta {\cos }^2\theta$
${\sin }^4\theta +{\cos }^4\theta =2{\sin }^2\theta {\cos }^2\theta$
(Use equation (2))
${\sin }^4\theta +{\cos }^4\theta =2\left(\frac{1}{4}\right)$
${\sin }^4\theta +{\cos }^4\theta =\frac{1}{2}$
Hence option (C) is correct.

Question:14

sin (45° + θ) – cos (45° – θ) is equal to
(A) 2cosθ (B) 0 (C) 2sinθ (D) 1

Answer:

Answer. [B]
Solution. Here
:$\sin ({45}^{\circ }+\theta )-\cos ({45}^{\circ }-\theta )$
Sin[90° - (45°- θ)] – cos(45°- θ)
$[\because ({45}^{\circ }+\theta )=({90}^{\circ }-(45-\theta ))]$
Cos(45°- θ) – cos(45°- θ) [$\because$ sin (90 – θ) = cosθ]
= 0
Hence option (B) is correct

Question:15

A pole 6 m high casts a shadow 2$\sqrt{3}$ m long on the ground, then the Sun’s elevation is
(A) 60° (B) 45° (C) 30° (D) 90°

Answer:

Answer. [A]
Solution. Given :
height pole = 6 m
Shadow of pole = $2\sqrt{3}m$
Now make figure according to given condition

Let angle of elevation is $\alpha$
$\therefore \tan \alpha =\frac{Perpendicular}{base}$
$\tan \alpha =\frac{6}{2\sqrt{3}}$
$\tan \alpha =\frac{3}{\sqrt{3}}=\sqrt{3}$
$\tan \alpha =\tan {60}^{\circ }$ $[\because \tan {60}^{\circ }=\sqrt{3}]$
a = 60°
Hence the Sun's elevation is 60°.

Question:1

Write ‘True’ or ‘False’ and justify your answer in each of the following:
$\frac{\tan {47}^{\circ }}{\cot {43}^{\circ }}=1$

Answer:

$\frac{\tan {47}^{\circ }}{\cot {43}^{\circ }}=1$
Taking L.H.S.
$\frac{\tan {47}^{\circ }}{\cot {43}^{\circ }}$
$\frac{\tan ({90}^{\circ }-{43}^{\circ })}{\cot {43}^{\circ }}$ $(\because 47=(90-43))$
$\frac{\cot {43}^{\circ }}{\cot {43}^{\circ }}=1$ $(\because \tan (90-\theta )=\cot \theta )$
Hence L.H.S. = R.H.S.
So, the given expression is true.

Question:2

The value of the expression (cos² 23° – sin² 67°) is positive.

Answer:

Answer. [False]
Solution. (cos² 23° – sin² 67°)
= $({\cos }^2({90}^{\circ }-{67}^{\circ })-{\sin }^2{67}^{\circ })$ $(\because {23}^{\circ }={90}^{\circ }-{67}^{\circ })$
= Sin²67° - sin²67° (cos(90-θ) = sin θ)
= 0
Hence the value of the expression is neutral
So, the given statement is false.

Question:3

The value of the expression (sin 80° – cos80°) is negative.

Answer:

Answer. [False]
Solution. (sin 80° – cos 80°)
We know that from 0 to 90° sin$\theta$ and cos$\theta$ both are positive i.e.
$0<\sin \theta \le {90}^{\circ }$ (always positive)
$0<\cos \theta \le {90}^{\circ }$ (always positive)
At 45° both the values of sin$\theta$ and cos$\theta$ are the same but after 45° to 90° value of sin is greater than the value of cos$\theta$ Hence sin 80° > cos 80°.
If we subtract a smaller term from bigger than the result is positive.
Hence (sin 80° – cos80°) > 0
So, the given statement is false

Question:4

Prove that $\sqrt{(1-{\cos }^2\theta ){\sec }^2\theta }=\tan \theta$

Answer:
L.H.S
$\sqrt{(1-{\cos }^2\theta ){\sec }^2\theta }\cdots \left(1\right)$
We know that
$1-{\cos }^2\theta ={\sin }^2\theta \text{ in (1)}$
$\sqrt{{\sin }^2\theta \cdot {\sec }^2\theta }$
$=\sqrt{{\sin }^2\theta \times \frac{1}{{\cos }^2\theta }}$ $(\because \sec \theta =\frac{1}{\cos \theta })$
$=\sqrt{\frac{{\sin }^2\theta }{{\cos }^2\theta }}$
$=\sqrt{{\tan }^2\theta }=\tan \theta$ $(\because \frac{\sin \theta }{\cos \theta }=\tan \theta )$
L.H.S. = R.H.S.
Hence the given expression is true.

Question:5

If cosA + cos²A = 1, then sin²A + sin⁴A = 1.

Answer:

Given
cosA + cos²A = 1 …(1) cos A = 1 – cos²A
cosA = sin²A …(2) ($\because$ sin²$\theta$ = 1 – cos²$\theta$)
${\sin }^{2}A+{\sin }^{4}A=1$
L.H.S.
${\sin }^{2}A+{\sin }^{4}A$
${\sin }^{2}A+{({\sin }^{2}A)}^2$
$\cos A+{(\cos A)}^2$ (from (2))
cosA + cos²A
= 1 (R.H.S.) (from (1))
Hence sin²A + sin⁴A = 1
So, the given statement is true.

Question:6

(tanθ + 2) (2 tan θ + 1) = 5 tan θ + sec²θ.

Answer:

(tanθ + 2) (2 tan θ + 1) = 5 tan θ + sec²θ
Taking L.H.S.
(tanθ + 2) (2 tan θ + 1)
tanθ.(2tan θ+1) + 2(2tan θ +1)
$2{\tan }^2\theta +\tan \theta +4\tan \theta +2$
$2{\tan }^2\theta +5\tan \theta +2$
$2({\tan }^2\theta +1)+5\tan \theta \cdots \left(1\right)$
We know that
${\sec }^2\theta -{\tan }^2\theta =1$
$(1+{\tan }^2\theta ={\sec }^2\theta )$
Put the above value in (1)we get
$2{\sec }^2\theta +5\tan \theta \ne 5\tan \theta +{\sec }^2\theta$
L.H.S. $\ne$ R.H.S.
Hence the given expression is false.

Question:7

If the length of the shadow of a tower is increasing, then the angle of elevation of the sun is also increasing.

Answer:

Answer. [False]
Solution. Let us take 2 cases.
Case 1:

Case 2 :

${\theta }_1$ is the angle when length is small and ${\theta }_2$ is the angle when shadow length is increased.
for finding ${\theta }_1$, ${\theta }_2$ find tan $\theta$
$\tan {\theta }_1=\frac{Perpendicular}{Base}=\frac{heightoftower}{lengthofshadow}$
$\tan {\theta }_2=\frac{heightoftower}{lengthofshadow}$
In both the case height of the tower is the same but in case 2 length of the shadow is increased and if the length of shadow increased value of ${\theta }_2$ decreased.
Hence the given statement is false.

Question:8

If a man standing on a platform 3 metres above the surface of a lake observes a cloud and its reflection in the lake, then the angle of elevation of the cloud is equal to the angle of depression of its reflection.

Answer:

Answer. [False]
Solution. According to question.

In the figure ${\theta }_1$ is the angle of elevation and ${\theta }_2$ is the angle of depression of the cloud.
For finding ${\theta }_1$find tanq in $△$ECD
$\tan {\theta }_1=\frac{perpendicular}{Base}=\frac{DC}{EC}=\frac{H}{\iota }$
Find tan$\theta$ in $△$ECB for ${\theta }_2$
$\tan {\theta }_2=\frac{BC}{EC}=\frac{3}{\iota }$
Here we found that ${\theta }_1$ and ${\theta }_2$ both are different.
Hence the given statement is false.

Question:9

The value of 2sinθ can be a+1/a , where a is a positive number, and a ≠ 1.

Answer:

Answer. [False]
Solution. We know that
-1≤ sin θ ≤ 1
Multiply by 2.
-2≤ 2 sin θ ≤ 2
Here we found that value of 2 sin $\theta$ is lies from – 2 to 2.
But if we take a > 0 and a $\ne$ 1 then
$a+\frac{1}{a}>2$
For example a = 3
3 + 1/3 = 3.33
Hence $a+\frac{1}{a}$ is always greater than 2 in case of positive number except 1
But value of 2 sin $\theta$ is not greater than 2
Hence the given statement is false

Question:10

$\cos \theta =\frac{a^2+b^2}{2ab}$ where a and b are two distinct numbers such that ab> 0.

Answer:

We know that
$-1\le \cos \theta \le 1$
We also know that
${(a-b)}^2=a^2+b^2-2ab$
$\text{Since }$ ${(a-b)}^2$ $\text{is a square term hence it is always positive }$
${(a-b)}^2>0$
a² + b² – 2ab > 0
$a^2+b^2>2ab$
We observe that $a^2+b^2$ is always greater than 2ab.
Hence,
$\frac{a^2+b^2}{2ab}>1$
Because if we divide a big term by small then the result is always greater than 1.
cos$\theta$ is always less than or equal to 1
Hence the given statement is false.

Question:11

The angle of elevation of the top of a tower is 30°. If the height of the tower is doubled, then the angle of elevation of its top will also be doubled.

Answer:

Answer. [False]
Solution. According to question

Case: 1
Here BC is the tower.
Let the height of the tower is H and distance AB = a
In $△$ABC
$\tan \theta =\frac{Perpendicular}{Base}$
$\tan {30}^{\circ }=\frac{H}{a}$ $(\because \theta ={30}^{\circ })$
$\frac{1}{\sqrt{3}}=\frac{H}{a}\cdots \left(1\right)$ $(\because \tan {30}^{\circ }=\frac{1}{\sqrt{3}})$
Case :2 When height is doubled

Here ED = a
In $△DEF$
$\tan \theta =\frac{2H}{a}$
$\tan \theta =\frac{2}{3}$ (from (1))
But $\tan {60}^{\circ }=\sqrt{3}$ (If the angle is double)
$\sqrt{3}\ne \frac{2}{\sqrt{3}}$
hence the given statement is false.

Question:12

If the height of a tower and the distance of the point of observation from its foot,both, are increased by 10%, then the angle of elevation of its top remains unchanged.

Answer:

Answer. [True]
Solution. According to question


In case-1. Height is H and observation distance is a.
In case-2, both height and observation distance is increased by 10%.
In case -1
$\tan {\theta }_1=\frac{H}{a}$ $(\because \tan \theta =\frac{perpendicular}{Base})$ .....(1)
In case -2
$\tan {\theta }_2=\frac{H+\frac{H}{10}}{a+\frac{a}{10}}$
$=\frac{\frac{11H}{10}}{\frac{11a}{10}}=\frac{11H}{10}\times \frac{10}{11a}=\frac{H}{a}$
$\tan {\theta }_2=\frac{H}{a}\cdots \left(2\right)$
from equation (1) and (2) we observe that ${\theta }_1={\theta }_2$
Hence the given statement is true.

Question:1

Prove the following
:$\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }=2\cos ec\theta$

Answer:

$\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }=2\cos ec\theta$
Taking L.H.S.
$=\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }$
Taking LCM
$=\frac{{\sin }^2\theta +{(1+\cos \theta )}^2}{(1+\cos \theta )\sin \theta }$
$=\frac{{\sin }^2\theta +{\cos }^2\theta +2\cos \theta }{(1+\cos \theta )\sin \theta }$ $(\because {(a+b)}^2=a^2+b^2+2ab)$
$=\frac{1+1+2\cos \theta }{(1+\cos \theta )\sin \theta }$ $(\because {\sin }^2\theta +{\cos }^2\theta =1)$
$=\frac{2(1+\cos \theta )}{(1+\cos \theta )\sin \theta }$
$=\frac{2}{\sin \theta }$
$=2\cos ec\theta$ $(\because \frac{1}{\sin \theta }=\cos ec\theta )$

L.H.S. = R.H.S.
Hence proved.

Question:2

Prove the following :
$\frac{tanA}{1+\sec A}-\frac{\tan A}{1-\sec A}=2cosecA$

Answer:

Solution.
$\frac{tanA}{1+\sec A}-\frac{\tan A}{1-\sec A}=2cosecA$
Taking L.H.S.
$=\frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}$
Taking L.C.M.
$=\frac{\tan A(1-\sec A)-\tan A(1+\sec A)}{(1+\sec A)(1-\sec A)}$
$$\begin{gathered} =\frac{\tan A-\tan A \\ secA-\tan A-\tan A\sec A}{1-{\sec }^{2}A} \end{gathered}$$ $(\because (a-b)(a+b)=a^2-b^2)$
$=\frac{-2\tan A\sec A}{-{\tan }^{2}A}$ $(\because {\sec }^{2}A-{\tan }^{2}A=1)$
$=\frac{2\sec A}{\tan A}$
$=\frac{2}{\cos A}\times \frac{\cos A}{\sin A}$ $\left(\begin{array}{cc}\because \sec \theta =\frac{1}{\cos \theta }& \\ \because \tan \theta =\frac{\sin \theta }{\cos \theta }& \end{array}\right)$
$=\frac{2}{\sin A}=2\cos ecA$ $(\because \frac{1}{\sin \theta }=\cos ec\theta )$
L.H.S. = R.H.S.
Hence proved

Question:3

If tan A = 3/4 , then show that sinAcos A = 12/25 .

Answer:

Given:-
$\tan A=\frac{3}{4}$
To prove:-
$sinA\cos A=\frac{12}{25}...........(1)$
we know that
$\tan \theta =\frac{P}{B}$
$\frac{P}{B}=\frac{3}{4}$
P=3, B = 4
Using Pythagoras theorem
$H^2=B^2+P^2$
$H^2=\left(4^2\right)+\left(3^2\right)$
$H^2=9+16$
$H^2=25$
$H=\sqrt{25}=5$
H = 5
we know that
$\sin \theta =\frac{P}{H},\cos \theta =\frac{B}{H}$
Hence
$\sin A=\frac{3}{5},\cos A=\frac{4}{5}$
Put the value of sinA and cosA in equation (1)
$\frac{3}{5}\times \frac{4}{5}=\frac{12}{25}$
L.H.S. = R.H.S.
Hence proved.

Question:4

Prove the following : (sin α + cos α) (tan α + cot α) = sec α + cosec α

Answer:

Solution.
(sin α + cos α) (tan α + cot α) = sec α + cosec α
Taking L.H.S.
$(\sin \alpha +\cos \alpha )(\tan \alpha +\cot \alpha )$
$=(\sin \alpha +\cos \alpha )(\frac{\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{\sin \alpha })(\because \tan \theta =\frac{\sin \theta }{\cos \theta },\cot \theta =\frac{\cos \theta }{\sin \theta })$
Taking L.C.M.
$=\frac{(\sin \alpha +\cos \alpha )({\sin }^2\alpha +{\cos }^2\alpha )}{\sin \alpha +\cos \alpha }$
$=\frac{\sin \alpha }{\sin \alpha \cos \alpha }+\frac{\cos \alpha }{\sin \alpha +\cos \alpha }$ $(\because {\sin }^2\theta +{\cos }^2\theta =1)$
by separately divide
$=\frac{\sin \alpha }{\sin \alpha \cos \alpha }+\frac{\cos \alpha }{\sin \alpha +\cos \alpha }$
$=\frac{1}{\cos \alpha }+\frac{1}{\sin \alpha }$
$=\sec \alpha +\cos ec\alpha$ $(\because \frac{1}{\cos \theta }=\sec \theta ,\frac{1}{\sin \theta }=\cos ec\theta )$
L.H.S. = R.H.S.
Hence proved.

Question:5

Prove the following :
$(\sqrt{3}+1)(3-\cot {30}^{\circ })={\tan }^3{60}^{\circ }-2\sin {60}^{\circ }$
Answer:

$(\sqrt{3}+1)(3-\cot {30}^{\circ })={\tan }^3{60}^{\circ }-2\sin {60}^{\circ }$
Taking L.H.S
$(\sqrt{3}+1)(3-\cot {30}^{\circ })$
$=\sqrt{3}(3-\cot {30}^{\circ })+1(3-\cot {30}^{\circ })$
$=3\sqrt{3}-\sqrt{3}\cot {30}^{\circ }+3-\cot {30}^{\circ }$
We know that
$\cot {30}^{\circ }=\sqrt{3}$
$=3\sqrt{3}-\sqrt{3}\left(\sqrt{3}\right)+3-\sqrt{3}$
$=3\sqrt{3}-3+3-\sqrt{3}$
$=3\sqrt{3}-\sqrt{3}$
$=2\sqrt{3}$
R.H.S.
${\tan }^3{60}^{\circ }-2\sin {60}^{\circ }$
We know that
$tan{60}^o=\sqrt{3}$
${\tan }^3{60}^{\circ }-2\sin {60}^{\circ }={\left(\sqrt{3}\right)}^3-2\left(\frac{\sqrt{3}}{2}\right)$
$=3\sqrt{3}-\sqrt{3}=2\sqrt{3}$
Hence proved

Question:6

Prove the following :
$1+\frac{{\cot }^2\alpha }{1+\cos ec\alpha }=\cos ec\alpha$

Answer:

$1+\frac{{\cot }^2\alpha }{1+\cos ec\alpha }=\cos ec\alpha$
Taking L.H.S.
$=1+\frac{{\cot }^2\alpha }{1+\cos ec\alpha }$

$=\frac{1+\cos ec\alpha +{\cot }^2\alpha }{1+\cos ec\alpha }$
$=\frac{\cos e{c}^2\alpha -{\cot }^2\alpha +\cos ec\alpha +{\cot }^2\alpha }{1+\cos ec\alpha }$ $(\because \cos e{c}^2\theta -{\cot }^2\theta =1)$
$=\frac{\cos e{c}^2\alpha +\cos ec\alpha }{1+\cos ec\alpha }$
$=\frac{\cos ec\alpha (\cos ec\alpha +1)}{(1+\cos ec\alpha )}$
$=\cos ec\alpha$
L.H.S. = R.H.S.
Hence proved.

Question:7

Prove the following: tan θ + tan (90° – θ) = sec θ sec (90° – θ)

Answer:

Solution.
tan θ + tan (90° – θ) = sec θ sec (90° – θ)
Taking L.H.S.
= tan θ + tan (90° – θ)
= tanθ + cotθ ($\because$ tan (90 – θ) = cot θ)
$=\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }$ $(\because \tan \theta =\frac{\sin \theta }{\cos \theta },\cot \theta =\frac{\cos \theta }{\sin \theta })$
Taking L.C.M.
$\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }$
$=\frac{1}{\cos \theta \cdot \sin \theta }$ $(\because {\sin }^2\theta +{\cos }^2\theta =1)$
$=\frac{1}{\cos \theta }\times \frac{1}{\sin \theta }$
$=\sec \theta \times \cos ec\theta$ $(\because \frac{1}{\cos \theta }=\sec \theta ,\frac{1}{\sin \theta }=\cos ec\theta )$
$=\sec \theta \times \sec ({90}^{\circ }-\theta )$ $(\because \sec (90-\theta )=\cos ec\theta )$
L.H.S. = R.H.S.
Hence proved.