If sin θ = 12/13, find the value of sin² θ- cos² θ/2 sin θ.cos θ x 1/tan² θ
Last Updated :
18 Feb, 2024
Trigonometry is basically the study of the relationship between the angles and the sides of a triangle. It is one of the widely used topics of Mathematics that is used in daily life. It involves operations on a right-angled triangle i.e. a triangle having one of the angles equal to 90°. There are some terms that we should know before going further. These terms are,
- Hypotenuse – it is the side opposite to the right angle in a right-angled triangle. It is the longest side of a right-angled triangle. In Figure 1, side AC is the hypotenuse.
- Perpendicular – the perpendicular of a triangle, corresponding to a particularly acute angle θ is the side opposite to the angle θ. In Figure 1, side AB is the perpendicular corresponding to angle θ.
- Base – it is the side adjacent to a particularly acute angle θ. In Figure 1 side BC is the base corresponding to angle θ.
Figure 1
As earlier said, trigonometry depicts the relationship between the angles and sides of a right-angled triangle. These relationships is represented by standard ratios and are given as follows,
- Sine (sin) The sine of an angle θ is the ratio of the length of the perpendicular, corresponding to the angle θ, to the length of the hypotenuse of the triangle.
sin θ = perpendicular/hypotenuse = p/h
- Cosine (cos) The cosine of an angle θ is the ratio of the length of the base, corresponding to the angle θ, to the length of the hypotenuse of the triangle.
cos θ = base/hypotenuse = b/h
- Tangent (tan) The tangent of an angle θ is the ratio of the length of the perpendicular, corresponding to the angle θ, to the length of the base for the particular angle of the triangle.
tan θ = perpendicular/base = p/b
- Cotangent (cot) It is the reciprocal of a tangent.
cot θ = 1/tan θ=base/perpendicular = b/p
- Secant (sec) It is the reciprocal of cosine.
sec θ =1/cos θ = hypotenuse/base = h/b
- Cosecant (cosec) It is the reciprocal of sine.
cosec θ = 1/sin θ = hypotenuse/perpendicular = h/p
There also exist relations between each of these ratios and some of them that we will be using are,
- tan θ = sin θ/ cos θ
- cot θ = cos θ/ sin θ
- sin² θ + cos² θ =1
If sin θ = 12/13, 0° < θ < 90°, find the value of sin² θ – cos² θ /2 sin θ × cos θ × 1/tan² θ
Solution:
Given,
sin θ =12/13
sin² θ = 144/169
It is known,
sin² θ + cos² θ = 1
cos² θ = 1 – sin² θ
cos θ = √(1 – sin² θ)
Here, sin θ = 12/13
Therefore,
cos θ = √(1 – (12/13)2)
cos θ = √(1 – 144/169)
cos θ = √((169 – 144)/169)
cos θ = √(25/169)
cos θ = 5/13
and cos² θ = 25/169
tan θ = sin θ/cos θ
tan θ = (12/13)/(5/13)
tan θ = (12/13) × (13/5)
tan θ = 12/5
tan2 θ = 144/25
With all these in our hands, now find the value of our equation
(sin² θ – cos² θ )/(2 sin θ × cos θ ) × 1/tan² θ
= (144/169 – 25/169)/(2 × 12/13 × 5/13) × 1/(144/25)
= (119/169) / (120/169) × (25/144)
= (119/169) × (169/120) × (25/144)
= 0.172
Therefore, the required answer of the given equation is 0.172.
Similar Problems
Question 1: If sin θ = 1/2 and cos ϕ = √3/2 then find the value of (tan θ + tan ϕ) /(1- tan θ × tan ϕ).
Solution:
sin θ = 1/2
sin² θ + cos² θ = 1
cos² θ = 1 – sin² θ
cos θ = √(1 – sin² θ)
cos θ = √(1 – 1/4)
cos θ = √3/2
tan θ = sin θ / cos θ
= (1/2)/(√3/2)
= 1/√3
Similarly,
sin ϕ = √(1 – cos2 ϕ)
= √(1 – 3/4)
= 1/2
tan ϕ = sin ϕ / cos ϕ
= (1/2)/(√3/2)
= 1/√3
Therefore,
(tan θ + tan ϕ) /(1- tan θ × tan ϕ) = (1/√3 + 1/√3)/( 1 -1/√3 × 1/√3)
= (2/√3)/(1 – 1/3)
= (2/√3)/(2/3)
= (2/√3) × (3/2)
= √3
Question 2: If 5 cos x – 12 sin x = 0 then find the value of (sin x + cos x)/(2 cos x – sin x).
Solution:
5 cos x – 12 sin x =0
5 cos x = 12 sin x
5 = 12 × sinx/cos x
5 = 12 tanx
tan x = 5/12
(sin x + cos x)/(2 cos x – sin x)
dividing the numerator and denominator by cos x,
(tan x + 1)/(2 – tan x)
= (5/12 + 1)/(2 – 5/12)
= (17/12)/(19/12)
= 17/19
Question 3: If a cos x + b sin x= t and a sin x – b cos x = u, then find sin x and cos x.
Solution:
Given,
a cos x + b sin x = t ⇢ (i)
a sin x – b cos x = u ⇢ (ii)
By, b × (i) + a × (ii),
ab cos x + b2 sin x + a2 sin x – ab cos x = bt + a
sin x (a2 + b2)= bt + au
sin x = (bt + au)/(a2 + b2)
Similarly,
By, a × (i) – b × (ii), we get
a2 cos x + ab sin x – ab sin x + b2 cos x = at – bu
cos x (a2 + b2) = at – bu
cos x = (at – bu)/(a2 + b2)
Question 4: In the right-angled triangle ABC, angle B = 90° and tan C = 1/2. If AC = 5, find the lengths of the side AB and BC.
Solution:
Given, tan C = 1/2
tan C = p/b = AB/BC
Therefore,
tan C = AB/BC =1/2
Let, AB and BC be k and 2k respectively.
By Pythagoras’ theorem,
AB2 + BC2 = AC2
k2 + (2k)2 = 52
5 k2= 25
k2= 5
k = √5
Therefore,
AB = k = √5 and
BC = 2k = 2√5
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