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# NCERT Solutions of Important Questions for Class 10 Maths  Chapter 8 Introduction to Trigonometry

NCERT Solutions of important questions from Class 10 Maths Chapter 8 Introduction To Trigonometry are given here straightforward and well-ordered clarifications. These answers for Introduction To Trigonometry are amazingly well known among Class 10 students for Math Introduction To Trigonometry Solutions convenient for rapidly finishing your homework and getting ready for tests. All questions and answers from the NCERT Book of Class 10 Maths Chapter 8 are given here to you . You will likewise adore the advertisement-free understanding on future study point’s NCERT Solutions of Important questions. All NCERT Solutions of Important Questions for Class 10 Maths are set up by specialists and are 100% precise.

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## NCERT Solutions for Class 10 Maths  Chapter 8 Introduction to Trigonometry

Exercise 8.1- Introduction to Trigonometry

Exercise 8.2 -Introduction to Trigonometry

Exercise 8.3-Introduction to Trigonometry

NCERT solutions of Important Questions-Introduction to Trigonometry

## NCERT Solutions of Important Questions for Class 10 Maths  Chapter 8 Introduction to Trigonometry

For doing all the trigonometric solutions you should have the following formulas in your brain.

$sin\Theta = \frac{perpendicular}{hypotenuse}, \; cos\Theta = \frac{base}{hypotenuse}$

$tan\Theta = \frac{perpendicular}{base},\; cot\Theta =\frac{base}{perpendicula}$

$sec\Theta = \frac{hypotenuse}{base},\; cosec\Theta =\frac{hypotenuse}{perpendicular}$

Point to be noticed that here perpendicular means side opposite to the angle and base means side adjacent to the angle, the hypotenuse is the longest side opposite to right angle as in the following fig.

$\fn_cm Hint:sinC=\frac{P}{H}=\frac{AB}{AC}$    ,     $\fn_cm sinA=\frac{P'}{H}=\frac{BC}{AC}$

$sin\Theta = \frac{1}{cosec\Theta }\, , cosec\Theta = \frac{1}{sin\Theta }$

$cos\Theta =\frac{1}{sec\Theta \, }\: ,sec\Theta =\frac{1}{cos\Theta \: }$

$tan\Theta =\frac{1}{cot\Theta \: },cot\Theta =\frac{1}{tan\Theta }$

$sin^{2}\Theta + cos^{2}\Theta =1$

$1 + tan^{2}\Theta = sec^{2}\Theta$

$1 + cot^{2}\Theta = cosec^{2}\Theta$

## NCERT Solutions of Important Questions for Class 10 Maths  Chapter 8 Introduction to Trigonometry

#### Example –

Express cosθ into all other trigonometric ratios.

Solution-

transformation into sinθ

The relationship between cosθ and sinθ is given as

sin²θ + cos²θ = 1

cos²θ = 1 – sin²θ

$cos\Theta = \sqrt{1-sin^{2}\Theta }$

Transformation into tanθ

There is no direct relation between cosθ and tanθ but we know the relationship between cosθ and secθ as following.

$cos\Theta = \frac{1}{sec\Theta }$…………(1)

Determining the value of secθ in terms of tanθ.

1 + tan²θ = sec²θ

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

Substituting the value of secθ in (1)

$cos\Theta = \frac{1}{\sqrt{1-tan^{2}\Theta }}$

Transformation into cotθ

We have got the cosθ in terms of tanθ

$cos\Theta = \frac{1}{\sqrt{1-tan^{2}\Theta }}$ ………(1)

We know

$tan\Theta = \frac{1}{cot\Theta }$………..(2)

Substituting  this value of  tanθ in (1)

$cos\Theta = \frac{1}{\sqrt{1 + \frac{1}{cot^{2}\Theta } }}$

$cos\Theta = \frac{cot\Theta }{\sqrt{cot^{2}\Theta +1}}$

Transformation into secθ

There is a direct relationship between cosθ and secθ

$cos\Theta = \frac{1}{sec\Theta }$

Transformation into cosecθ

While expressing cosθ into cosecθ, think over the following fact

$Relationship \; between \; cos\Theta \; and \; sin\Theta \; and \; sin\Theta \; to \; cosec\Theta$

$cos\Theta = \sqrt{1-sin^{2}\Theta }$………..(1)

$sin\Theta = \frac{1}{cosec\Theta }$

Substituting the value of sinθ in (1)

$cos\Theta = \sqrt{1 - \frac{1}{cosec^{2}\Theta }}$

$cos\Theta = \frac{cosec\Theta }{\sqrt{cosec^{2}\Theta -1}}$

## NCERT Solutions of Important Questions for Class 10 Maths  Chapter 8 Introduction to Trigonometry

### Practice exercise-

(1) express sinθ in terms of tanθ.

(2) Write tan θ in phrases of cosecθ.

(3) decide the relationship between secθ and sinθ.

(4) remodel cosecθ into all trigonometric ratios.

(5)express cotθ in terms of secθ.

(1) Hint: sinθ⇒cosecθ⇒cotθ⇒tanθ

(2)Hint:tanθ⇒cotθ⇒cosecθ

(3)Hint:secθ⇒cosθ⇒sinθ

(4)Hint: Check the relationship between cosecθ to other trigonometric ratios one by one.

(5) Hint:cotθ⇒tanθ⇒secθ

#### Trigonometric Ratios of Complementary Angles-

∠A + ∠B + ∠C = 18O°

∠A + 90° + ∠C =180°

∠A +  ∠C = 180°– 90°

∠A +  ∠C = 90°

∠A & ∠C are complementry to each other

∠C = 90 –∠A

So, comparing the trigonometric ratios corresponding to ∠A, and ∠(90 –A)i.e∠C.

$sinA = \frac{BC}{AC},\; cosA = \frac{AB}{AC},\; tanA =\frac{BC}{AB}$

$cosecA = \frac{AC}{BC},\; secA = \frac{AC}{AB},\; cotA =\frac{AB}{BC}$……………(1)

$sin(90-A) = \frac{AB}{AC},\; cos(90-A) = \frac{BC}{AC},\; tan(90-A) =\frac{AB}{BC}$

$cosec(90-A) = \frac{AC}{AB},\; sec(90-A) = \frac{AC}{BC},\; cot(90-A) =\frac{BC}{AB}$……(2)

From the equations (1) and (2) we have

sin(90–A) = cosA, cos (90– A) = sinA, tan(90–A) = cotA

cot(90–A) = tanA, sec (90– A) = cosecA, cosec(90–A) = sinA

Determining the value of other trigonometric ratio when value of one of them is given-

Table of trigonometric functions for cbse 10 th class

Study NCERT solutions for class 10 maths exercise 8.1 Introduction to trigonometry

Example-

Find the value of tanθ if

${\color{DarkBlue} sin\Theta =\frac{7}{25}}$.

${\color{DarkBlue} sin\Theta =\frac{perpedicular}{hypotenuse}}$

sinθ = 7 : 25

Let perpendicular= 7x and hypotenuse= 25x

${\color{DarkBlue} base= \sqrt{h^{2}-p^{2}}}$

${\color{DarkBlue} \therefore base = \sqrt{\left ( 25x \right )^{2}-\left ( 7x \right )^{2}}}$

${\color{DarkBlue} base = \sqrt{576}x^{2} = 24x}$

${\color{DarkBlue} \because tan\Theta = \frac{p}{b}}$

${\color{DarkBlue} \therefore \\ tan\Theta = \frac{7x}{24x}}$

${\color{DarkBlue} \\ tan\Theta = \frac{7}{24}}$

Ans.

## NCERT Solutions of Important Questions for Class 10 Maths  Chapter 8 Introduction to Trigonometry

Practice exercise- Evaluate the following.

$(1)){\color{DarkBlue} \mathbf{\frac{5cos^{2}60^{\circ}+4sec^{2}30^{\circ}-tan^{2}45}{sin^{2}30^{\circ}+cos^{2}30^{\circ}}}}$

${\color{DarkBlue} \mathbf{(2)}\mathbf{sin67^{\circ}cos23^{\circ}+sin23^{\circ}cos67^{\circ}}}$

(3)   In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

${\color{DarkBlue} \mathbf{(4)\; tan\left ( \frac{A+B}{2} \right )=cot\frac{C}{2}}}$

Study A to Z NCERT 10 th class maths solution of lesson 9.1 ‘Some Application Of the Trigonometry’

${\color{DarkBlue} \mathbf{(i)\; \frac{sin^{2}67+sin^{2}23}{cos^{2}55+cos^{2}25}}}$

67° + 23° = 90° and 55° +25° = 90°⇒23= 90 –67 and 25 =90–55, so substituting th values of 23 and 25 in the given expression

${\color{DarkBlue} \mathbf{=\frac{sin^{2}67+sin^{2}\left ( 90-67 \right )}{cos^{2}55+cos^{2}\left ( 90-55 \right )}}}$

${\color{DarkBlue} \mathbf{=\frac{sin^{2}67+cos^{2} 67 }{cos^{2}55+sin^{2} 55 }}}$

${\color{DarkBlue} \mathbf{=\frac{1}{1}}}={\color{DarkBlue} \mathbf{} 1}$(sin²θ +cos²θ=1)

(2) If ∠A and ∠B are acute angles such that cosA = cosB then show that ∠A = ∠B.

Answer- Let ΔABC is a right triangle in which ∠A and ∠B are acute angles.

We are given that cosA = cosB

$\fn_cm cosA=\frac{AC}{AB}$

$\fn_cm cosB=\frac{BC}{AB}$

$\fn_cm \frac{AC}{AB}=\frac{BC}{AB}$

AC = BC

∠B = ∠A(angles to equal opposite sides)

See the video for solutions of Q5 part(i),part(ii) and part(iii)  of exercise 8.4 class 10 NCERT maths

$\mathbf{{\color{DarkBlue} }(3)}\mathbf{{\color{DarkBlue} \left ( cosec\Theta -cot\Theta \right )^{2}= \frac{1-cos\Theta }{1+cos\Theta }}}$

L.H.S

${\color{DarkBlue} \mathbf{=\left ( \frac{1}{sin\Theta }-\frac{cos\Theta }{sin\Theta } \right )^{2}}}$

${\color{DarkBlue} \mathbf{=\left ( \frac{1-cos\Theta }{sin\Theta } \right )^{2}}}$

$=\mathbf{{\color{DarkBlue} \mathbf{}{\mathbf{ }}\frac{\left ( 1-cos\Theta \right )^{2}}{sin^{2}\Theta }}}$

Substituting sin²θ = 1 – cos²θ

$=\mathbf{{\color{DarkBlue} \mathbf{}{\mathbf{ }}\frac{\left ( 1-cos\Theta \right )^{2}}{1-cos^{2}\Theta }}}$

$=\mathbf{{\color{DarkBlue} \mathbf{}{\mathbf{ }}\frac{\left ( 1-cos\Theta \right )^{2}}{\left (1-cos\Theta \right )\left ( 1+cos\Theta \right ) }}}$

${\color{DarkBlue} \mathbf{{\color{DarkBlue} }\mathbf{ }=\frac{1-cos\Theta }{1+cos\Theta }}}$

= R.H.S

${\color{DarkBlue} (4)}{\color{DarkBlue} \mathbf{\frac{sin\Theta }{1+cos\Theta }+ \frac{1+cos\Theta }{sin\Theta }= 2cosec\Theta}}$

Solution-

${\color{DarkBlue} }{\color{DarkBlue} \mathbf{\frac{sin\Theta }{1+cos\Theta }+ \frac{1+cos\Theta }{sin\Theta }= 2cosec\Theta}}$

L.H.S

${\color{DarkBlue} \mathbf{=\frac{sin^{2}\Theta +\left ( 1+cos\Theta \right )^{2}}{sin\Theta \left ( 1+cos\Theta \right )}}}$

${\color{DarkBlue} =}{\color{DarkBlue} \mathbf{\frac{sin^{2}\Theta + 1 +cos^{2}\Theta +2cos\Theta }{sin\Theta \left ( 1+cos\Theta \right )}}}$

${\color{DarkBlue} =}{\color{DarkBlue} \mathbf{\frac{1 + 1 + 2cos\Theta }{sin\Theta \left ( 1+cos\Theta \right )}}}$

${\color{DarkBlue} =}{\color{DarkBlue} \mathbf{\frac{2\left ( 1+cos\Theta \right ) }{sin\Theta \left ( 1+cos\Theta \right )}}}$

${\color{DarkBlue} \mathbf{=\frac{2}{sin\Theta } }}$

= 2cosecθ= R.H.S

(5) Prove that ; (sinα + cosα)(tanα + cotα) = (secα + cosecα)

Solution-

L.H.S

(sinα + cosα)(tanα + cotα)

$\fn_cm =\left ( sin\alpha +cos\alpha \right )\left ( \frac{sin\alpha }{cos\alpha }+\frac{cos\alpha }{sin\alpha } \right )$

$\mathbf{={\color{DarkBlue} \left ( cos\alpha +sin\alpha \right )\left (\frac{sin^{2}\alpha +cos^{2}\alpha }{sin\alpha .cos\alpha } \right )}}$

$\mathbf{={\color{DarkBlue} \left ( cos\alpha +sin\alpha \right )\left (\frac{1}{sin\alpha .cos\alpha } \right )}}$

$\mathbf{{\color{DarkBlue} =\frac{cos\alpha +sin\alpha }{sin\alpha .cos\alpha }}}$

${\color{DarkBlue} \mathbf{\mathbf{{\color{DarkBlue} =\frac{cos\alpha }{sin\alpha .cos\alpha }}}+\frac{sin\alpha }{sin\alpha .cos\alpha }}}$

${\color{DarkBlue} \mathbf{\mathbf{{\color{DarkBlue} =\frac{1 }{sin\alpha }}}+\frac{1 }{ cos\alpha }}}$

= cosecα + secα= R.H.S

(6) Prove that

${\color{DarkBlue} \mathbf{\frac{cotA-cosA}{cotA +cosA}= \frac{cosecA-1}{cosecA +1}}}$

Solution-

L.H.S

${\color{DarkBlue} \mathbf{\frac{cotA-cosA}{cotA+ cosA}}}$

${\color{DarkBlue} \mathbf{=\frac{\frac{cosA}{sinA}-cosA}{\frac{cosA}{sinA}+cosA}}}$

${\color{DarkBlue} \mathbf{\mathbf{}=\frac{cosA\left [ \frac{1}{sinA}-1 \right ]}{cosA\left [ \frac{1}{sinA} -1\right ]}}}$

${\color{DarkBlue} \mathbf{=\frac{cosecA-1}{cosecA+1}}}$

= R.H.S

(7) Prove that

${\color{DarkBlue} \mathbf{\frac{sin\Theta -cos\Theta +1}{sin\Theta +cos\Theta -1}=\frac{1}{sec\Theta -tan\Theta }}}$

Since in the R.H.S we have to get secθ and tanθ so transforming L.H.S in terms of secθ and tanθ, dividing the numerator and denominator by cosθ, we get

${\color{DarkBlue} \mathbf{L.H.S=\frac{sin\Theta -cos\Theta +1}{sin\Theta +cos\Theta -1}=\frac{tan\Theta -1+sec\Theta }{tan\Theta +1-sec\Theta }}}$

${\color{DarkBlue} \mathbf{=\frac{\left ( tan\Theta +sec\Theta \right )-1}{\left ( tan\Theta -sec\Theta \right )+1}}}$

${\color{DarkBlue} \mathbf{{\color{DarkBlue} \mathbf{=\frac{\left [( tan\Theta +sec\Theta \right )-1]}{\left [( tan\Theta -sec\Theta \right )+1]}}}\frac{[tan\Theta -sec\Theta] }{[tan\Theta -sec\Theta] }}}$

Let the [tanθ – secθ] remain in its place at the denominator since in the R.H.S we are to be required this one.

So, multiplying the expressions in the numerators

${\color{DarkBlue} \mathbf{=\frac{\left ( tan^{2}\Theta - sec^{2}\Theta \right )-\left ( tan\Theta -sec\Theta \right )}{\left ( tan\Theta -sec\Theta +1 \right )\left ( tan\Theta -sec\Theta \right )}}}$

${\color{DarkBlue} \mathbf{=\frac{sec^{2}\Theta -1-sec^{2}\Theta -tan\Theta +sec\Theta }{\left ( tan\Theta -sec\Theta +1 \right )\left ( tan\Theta -sec\Theta \right )}}}$

${\color{DarkBlue} \mathbf{=\frac{-1-tan\Theta +sec\Theta }{\left ( tan\Theta -sec\Theta +1 \right )\left ( tan\Theta -sec\Theta \right )}}}$

${\color{DarkBlue} \mathbf{=\frac{-\left ( 1+tan\Theta -sec\Theta \right )}{\left ( tan\Theta - sec\Theta +1\right )\left ( tan\Theta -sec\Theta \right )}}}$

${\color{DarkBlue} \mathbf{=\frac{-1}{tan\Theta -sec\Theta }= \frac{1}{sec\Theta -tan\Theta }}}$

= R.H.S

(8) Prove that:

${\color{DarkBlue} \mathbf{\frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}=2secA}}$

L.H.S

${\color{DarkBlue} \mathbf{\frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}}}$

${\color{DarkBlue} \mathbf{=\frac{cos^{2}A+\left ( 1+sinA \right )^{2}}{cosA\left ( 1+sinA \right )}}}$

${\color{DarkBlue} \mathbf{=\frac{cos^{2}A+1+sin^{2}A+2sinA}{cosA\left (1+sinA \right )}}}$

${\color{DarkBlue} \mathbf{=\frac{1+1+2sinA}{cosA\left (1+sinA \right )}}}$

${\color{DarkBlue} \mathbf{=\frac{2\left ( 1+sinA \right )}{cosA\left (1+sinA \right )}}}$

${\color{DarkBlue} \mathbf{=\frac{2}{cosA}}}$

= 2secA = R.H.S

(9)

${\color{DarkBlue} \mathbf{\frac{tan\Theta }{1-cot\Theta }+\frac{cot\Theta }{1-tan\Theta }=1+ sec\Theta .cosec\Theta}}$

Solution-

L.H.S

$\dpi{100} {\color{DarkBlue} \mathbf{\frac{tan\Theta }{1-cot\Theta }+\frac{cot\Theta }{1-tan\Theta }}}$

$={\color{DarkBlue} \mathbf{\frac{\frac{sin\Theta }{cos\Theta } }{1-\frac{cos\Theta }{sin\Theta } }+\frac{\frac{cos\Theta }{sin\Theta } }{1-\frac{sin\Theta }{cos\Theta } }} }$

${\color{DarkBlue} \mathbf{=\frac{\frac{sin\Theta }{cos\Theta }}{\frac{sin\Theta -cos\Theta }{sin\Theta }}+\frac{\frac{cos\Theta }{sin\Theta }}{\frac{cos\Theta -sin\Theta }{cos\Theta }}}}$

$\dpi{100} \small {\color{DarkBlue} \mathbf{=\frac{sin^{2}\Theta }{cos\Theta \left ( sin\Theta -cos\Theta \right )}+\frac{cos^{2}\Theta }{sin\Theta \left ( cos\Theta -sin\Theta \right )}}}$

$\dpi{100} \small {\color{DarkBlue} \mathbf{=\frac{sin^{2}\Theta }{cos\Theta \left ( sin\Theta -cos\Theta \right )}-\frac{cos^{2}\Theta }{sin\Theta \left ( -cos\Theta +sin\Theta \right )}}}$

${\color{DarkBlue} \mathbf{=\frac{sin^{3}\Theta -cos^{3}\Theta }{sin\Theta .cos\Theta \left ( sin\Theta -cos\Theta \right )}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{\left ( sin\Theta -cos\Theta \right )\left ( sin^{2}\Theta +cos^{2}\Theta +sin\Theta .cos\Theta \right )}{sin\Theta .cos\Theta \left ( sin\Theta -cos\Theta \right )}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{\left ( sin\Theta -cos\Theta \right )\left ( 1 +sin\Theta .cos\Theta \right )}{sin\Theta .cos\Theta \left ( sin\Theta -cos\Theta \right )}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{1+sin\Theta .cos\Theta }{sin\Theta .cos\Theta }}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{1}{sin\Theta .cos\Theta }+ \frac{sin\Theta .cos\Theta }{sin\Theta .cos\Theta }}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{1}{sin\Theta }.\frac{1}{cos\Theta }+1}}$

=1 + secθ.cosecθ= R.H.S

(10) Prove that:

$\dpi{100} {\color{DarkBlue} \mathbf{\sqrt{\frac{1+sinA}{1-sinA}}= secA + tanA}}$

L.H.S

$\dpi{100} {\color{DarkBlue} \mathbf{\sqrt{\frac{1+sinA}{1-sinA}}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\sqrt{\frac{\left ( 1+sinA \right )\left ( 1+sinA \right )}{\left ( 1-sinA \right )\left ( 1+sinA \right )}}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{1+sinA}{\sqrt{1- sin^{2}A}}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{1+sinA}{\sqrt{cos^{2}A}}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{1+sinA}{cosA}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{1}{cosA} +\frac{sinA}{cosA}}}$

= secA + tanA = R.H.S

(11) Prove that:

$\fn_cm \left ( cosec\alpha -sinA \right )\left ( secA-cosecA \right )=\frac{1}{tanA+cotA}$

L.H.S

${\color{DarkBlue} \mathbf{{\color{DarkBlue} }\left ( coscA-sinA \right )\left ( secA- cosA \right )}}$

${\color{DarkBlue} \mathbf{=\left ( \frac{1}{sinA}-sinA \right )\left (cosA + \frac{1}{cosA} \right )}}$

$={\color{DarkBlue} \mathbf{\left ( \frac{1-sin^{2}A}{sinA} \right )\left ( \frac{1-cos^{2}A}{cosA} \right )= \frac{cos^{2}A.sin^{2}A}{sinA.cosA}}}$

${\color{DarkBlue} \mathbf{= \frac{sinA.cosA\left ( tanA +cotA \right )}{tanA+cotA}}}$

${\color{DarkBlue} \mathbf{=\frac{sinA.cosA\left ( \frac{sinA}{cosA}+\frac{cosA}{sinA} \right )}{tanA+cotA}}}$

${\color{DarkBlue} \mathbf{=\frac{sinA.cosA\left ( sin^{2}A+cos^{2}A \right )}{sinA.cosA}}}$

${\color{DarkBlue} \mathbf{=\frac{sinA.cosA\left ( sin^{2}A+cos^{2}A \right )}{sinA.cosA\left ( tanA+cotA \right )}}}$

${\color{DarkBlue} \mathbf{=\frac{1}{tanA+cotA}}}$

${\color{DarkBlue} \mathbf{{\color{DarkBlue} \mathbf{=\frac{1}{tanA+cotA}}}= R.H.S}}$

## NCERT Solutions of Science and Maths for Class 9,10,11 and 12

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