Class 10 Trignometry

Class 10 Trignometry

by Qaulity Education

​Topics covered

• ​Basic trignometry

• ​Trignometry ratios of common angles

• ​Trignometry ratios of complement angle

• ​Trignometry identities

• ​Question

​Some basic things to remember-

​Trignometry can only apply to right angle triangle.

1. ​Height is side taken opposite to angle taken.

2. ​Base is side on which angle taken lie,

3. ​Hypotenuse is the longest side of a right-angled triangle.

4. ​Sin,cos ratios can never be an improper fraction i.e. denominator is always larger than numerator.

5. ​Sec,cosec can never be a proper fraction.

6. ​Learn the ratios in order Sin,Cos,Tan.

Basic trignometry ratios

​Trick to remember ratios

​1st ratio is of sin that is L/K or P/H remember the order of ratios and trick

​Q1. Find all trignometric ratios in fig.

Multiple Choice

Find trignometry ratios sin x,cos x,tan x in following figure

• $$\frac{4}{\sqrt[]{65}}$$  , $$\frac{7}{\sqrt[]{65}}$$  , $$\frac{7}{4}$$  

• $$\frac{7}{\sqrt[]{65}}$$  , $$\frac{4}{\sqrt[]{65}}$$ , $$\frac{4}{7}$$   

• $$\frac{4}{\sqrt[]{65}}$$  , $$\frac{7}{\sqrt[]{65}}$$  , $$\frac{4}{7}$$  

• None of these

Multiple Choice

find tan P – cot R.

• 0

• -119/60

• 119/60

• 60/119

​Prove that tan x = sinx/cosx

​Hence proved.

Homework

​​Prove that cot x = cos x/sin x

​Ex 8.1 NCERT.

​Trignometric ratios of common angle(remember)

​Trick to remember table

1. ​Learn the angles in order 0,30,45,60,90.

2. ​Now as per order take the first line as sin.

3. Now write 0,1,2,3,4 in order beneath angles.​

4. ​Now divide the raw in 4 i.e. 0, 1/4 , 2/4 , 3/4 , 4/4

5. ​Than take square root of whole raw i.e. 0 , 1/2 , ​​​

​6. This is the first raw that is sin function at diffrent angles.

7. ​Now write inverse of first raw to find cos.

​8. Now divide first and second raw for tan.

​9. And the other threes are reciprocal of first three so basically reciprocal

​Q.1 Find sin 45 + cos 45.

​Q2.Find tan 60-cot 30.

​Do NCERT ex 8.2

Multiple Choice

$$\frac{\sin30\cos^230-\tan^260\sin30}{\operatorname{cosec}^260-\sec^{2\ }0}$$  =

• 16

• 0

• -9

• 9

Multiple Choice

$$\frac{\cos30-\sin\ 30}{\tan\ 45\ -\ \tan\ 60}$$  =

• $$\frac{\sqrt[]{3}}{2}$$  

• $$\sqrt[]{\frac{3}{2}}$$  

• $$\frac{2}{\sqrt[]{3}}$$  

• 0

​Trignometric ratios of complement angle

​Similiarly find all other

ratios

​Do ex. 8.3 NCERT

Multiple Choice

tan 48° tan 23° tan 42° tan 67° =

• 0

• UNDEFINED

• NONE

• 1

Multiple Choice

tan 2A = cot (A – 18°),find A

• -18

• 36

• 18

• 30/60

​Trignometric Identities (most important)

​Prove

​Similiarly divide eq. with

​ for II eq. and for III

​eq. to prove it.

Multiple Choice

$$9\sec^2A-9\tan^2A$$  

• cos A

• 9

• 1

• 0

Multiple Choice

(1+tan x+sec x)(1+cot x - cosec x)

• 2

• 2 cos x sin x

• 0

• 1

​Do exercise 8.4 or refer youtube,it is not possible to solve it here.

​If you want me to solve 8.4 or any queries in any question. You can mail me at [e-mail]

​THANK YOU