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# CBSE Class 10 Maths - Chapter 8 - Trigonometry

CONTENT LIST

## Introduction:

Trigonometry is the branch of mathematics' that focus on relationships of sides and angles of a triangle.

Let there be a right triangle ABC, right angled at C. Here ∠A (i.e. ∠CAB) is an acute angle, AB is hypotenuse, side BC is opposite to ∠A (called perpendicular) and side AC is adjacent to ∠A (called Base).

## Relational ship  between angle and side is expressed in termed of trigonometrical ratio.

Trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides.
The values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.

sine ∠A= Opposite/Hypotenuse = BC/AB= Perpendicular/ Hypotenuse

cosine ∠A= Adjacent / Hypotenuse = BC/AB= Base/ Hypotenuse

tangent ∠A= Opposite/Adjacent = BC/AC= Perpendicular/ Base

cosecant ∠A = Hypotenuse /Opposite= AB/BC =  Hypotenuse/Perpendicular/

secant ∠A = Hypotenuse/Adjacent   = AB/BC = Hypotenuse/ Base

cotangent ∠A = Adjacent /Opposite = AC/ BC =  Base/ Perpendicular

The all above trigonometric ratios are abbreviated as sin A, cos A, tan A, cosec A, sec A and cot A respectively. Trigonometric ratios are  also abbreviated as t-ratios.

If we write ∠A = θ, then the above results are as below:

sin θ = p/h  => reciprocal=> cosec θ = h/p

cos θ = b/h => reciprocal=> sec θ = h/b

tan θ = p/b => reciprocal=> cosec θ = b/p

## Trigonometrical Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

## Problem based on Trigonometry Class 10

Question 1.
If x =r sin θ and y =r cos θ then, the find value of x² + y² ?

Solution :

x² + y²
=>(rsinθ)² + (rcosθ)²
=>r²sin²θ + r²cos²θ
=>r²(sin²θ +cos²θ)
=>r²
Question 2.
Find the value of cosec70°-sec20° ?

Solution :

cosec70°-sec20°
=>cosec70°-sec20°
=>cosec70°-sec(90°-70°)
=>cosec70°-cosec70°=0
Question 3.
If 3 secθ - 5=θ then find the value cotθ ?

Solution :

Given,3 secθ - 5=0
=>secθ=5/3
=>cosθ=3/5
we know that,
sin²θ=1-cos²θ=1-(3/5)²
=>sin²θ=1-9/25
=>sinθ=4/5
∴cotθ=cos0/sinθ=3/5x5/4=3/5
Question 4.
If θ = 45°, then find value of sec θ cot θ - cosec θ tan θ ?

Solution :

sec θ cot θ - cosec θ tan θ
=>sec45°cot45° - cosec45° tan45°
=> √2X1-√2X1=0
Question 5.
If sin(90° - θ)cosθ= 1 and, θ is an acute angle then find the vlaue of θ ?

Solution :

sin(90° - θ)cosθ= 1
=>cosθ.cosθ= 1
=>cos²θ=1
=>cos²θ=cos²0°
=>θ=0°
Question 6.
If Triangle ΔTRY is a right angled isosceles triangle then, find the value of cosT +cos R +cos Y ?

Solution :

Let angle ∠R is 90°.
TR and RY are two equal sides of ΔTRY.
The sum of three angles is 180°.
We know that
TRY+RYT+RTY=180°
∠TRY = 90°.
∠RYT=∠RTY= x (say)
Value of cos T + cos R + cos Y will be-
=>cos 45° + cos 90° + cos 45 °
=>1/√2+0+1/√2
=>√2
Question 7.
If Δ ABC & Δ PRT are similar such that ∠C = ∠R =90° and AC/AB =3/5 then Find the value of sinT ?

Solution :

Δ ABC & Δ PRT are similar.
∠C = ∠R = 90°
Since the triangles are similar,
∠B = ∠ T
=> sin B = sin T (when two angles are equal, their sine values are also equal).
Now, in Δ ABC,
=>AC/AB = 3/5
=>sin B= AC/AB
=>sin B= 3/5
Therefore, sin T = sin B = 3/5
Question 8.
If k+7sec²62° -7 cot²28°=7sec 0°then, the Find the value of k ?

Solution :

k+7sec²62° -7 cot²28°=7sec 0°
=>k+7sec²62° - 7 Cot² (90 - 62)° = 7sec 0°
=> k + 7(sec²62° - Tan²62°) = 7 Sec0° (∵Sec²θ - Tan²θ = 1 and Sec0° = 1)
=>k + 7 = 7
=>k = 0
Question 9.
Sin²20° + sin² 70°/2 (cos²69° + cos² 21°) =sec60/k then Determine the value of k ?

Solution :

L.H.S
=>[sin²20° + sin²70°] / 2(cos²69° + cos²21°)
[We know that sinθ = cos(90-θ); cosθ = sin(90 - θ)]
sin70° = cos(90-70) = cos20°
cos69° = sin(90-69) = sin21°

=> [sin²20° + sin²70°] / 2(cos²69° + cos²21°)
=> [sin²20° + cos²20°] / 2 (sin²21° + cos²21°)
=> 1/2
R.H.S = L.H.S
=>Sec60/k = 1/2
=>2/k = 1/2
=>k = 4.
Question 10.
If sin (A - B) = 1/2 and cos (A + B) = 1/2, then Find the value of B ?

Solution :

Given,sin (A - B) = 1/2 and cos (A + B) = 1/2
=>sin (A - B) = 1/2
=>sin (A - B) = sin 30°
(A - B) = 30° ---------- eqn 1
=>cos (A + B) = 1/2
=>cos (A + B) = cos 60°
=>(A + B) = 60° ----------eqn 2
Adding eqn 1 and eqn 2 We get,
=> 2A=90°
=> A=45°
Putting A=45° in equation 1 we get,
=> B= 15°
Question 11.
If sin θ + sin² θ = 1, then find the value of cos² θ + (cos² θ)² ?

Solution :

Given,sin θ + sin² θ = 1
=>sin θ + sin² θ = 1
=>sinθ = 1 - sin² θ
=>sinθ = cos² θ
=>cos² θ + (cos² θ)²
=>sin θ + sin² θ
=>1 (given in quetion)
Question 12.
If tan x + sin x = m and tan x - sin x = n, then find the value of (m² - n²) in terms of m and n?

Solution :

Given, m = tan x + sin x and n = tan x - sin x

m² = (tan x + sin x)² = tan²x + sin²x + 2.tan x.sin x
n² = (tan x - sin x)² = tan²x + sin²x - 2.tan x.sin x

=> m² - n² = (tan²x + sin²x + 2.tan x.sin x) - (tan²x + sin²x - 2.tan x.sin x)
=> m² - n² = tan²x + sin²x + 2.tan x.sin x - tan²x - sin²x + 2.tan x.sin x
=> m² - n² = 4.tan x.sin x

=> m² - n² = 4√(tan²x.sin²x)
=> m² - n² = 4√((sin²x/cos²x).sin²x)
=> m² - n² = 4√(((1 - cos²x)/cos²x).sin²x)
=> m² - n² = 4√(((sin²x - sin²x.cos²x)/cos²x))
=> m² - n² = 4√(tan²x - sin²x)
=> m² - n² = 4√(tan x - sin x)(tan x + sin x)
=> m² - n² = 4√mn