Trigonometry is the branch of mathematics which deals with measurements of angles and the problems related to relationship between sides and angles.
Also, by Pythagoras theorem, we know that h^{2} = b^{2 }+ p^{2}.
Here, θ is pronounced as theta, ∠A=θ. We write trigonometric ratios according to angle θ.
If one of the angles of a rightangled triangle is 45°, then another angle will also be equal to 45°.
Let us say ABC is a rightangled triangle at B, such that;
∠ A = ∠ C = 45°
Thus, BC = AB = a (say)
Using Pythagoras theorem, we have;
AC^{2} = AB^{2} + BC^{2}
= a^{2} + a^{2}
= 2a^{2}
AC = a√2
Now, from the trigonometric ratios, we have;
sin 45° = (Opp. side to angle 45°)/Hypotenuse = BC/AC = a/a√2 = 1/√2
cos 45° = (Adj. side to angle 45°)/Hypotenuse = AB/AC = a/a√2 = 1/√2
tan 45° = BC/AB = a/a = 1
Similarly,
cosec 45° = 1/sin 45° = √2
sec 45° = 1/cos 45° = √2
cot 45° = 1/tan 45° = 1
Here, we will consider an equilateral triangle ABC, such that;
AB = BC = AC = 2a
∠A = ∠B = ∠C = 60°
Now, draw a perpendicular AD from vertex A that meets BC at D
According to the congruency of the triangle, we can say;
Δ ABD ≅ Δ ACD
Hence,
BD = DC
∠ BAD = ∠ CAD (By CPCT)
Now, in triangle ABD, ∠ BAD = 30° and ∠ ABD = 60°
Using Pythagoras theorem
AD^{2} = AB^{2} – BD^{2}
= (2a)^{2} – (a)^{2}
= 3a
AD = a√3
So, the trigonometric ratios for a 30degree angle will be;
sin 30° = BD/AB = a/2a = 1/2
cos 30° = AD/AB = a√3/2a = √3/2
tan 30° = BD/AD = a/a√3 = 1/√3
Also,
cosec 30° = 1/sin 30 = 2
sec 30° = 1/cos 30 = 2/√3
cot 30° = 1/tan 30 = √3
Similarly, we can derive the values of trigonometric ratios for 60°.
sin 60° = √3/2
cos 60° = 1/2
tan 60° = √3
cosec 60° = 2/√3
sec 60° = 2
cot 60° = 1/√3
If ABC is a rightangled triangle at B, if ∠A is reduced, then side AC will come near to side AB. So, if ∠ A is nearing 0 degree, then AC becomes almost equal to AB, and BC gets almost equal to 0.
Hence, Sin A = BC /AC = 0
and cos A = AB/AC = 1
tan A = sin A/cos A = 0/1 = 0
Also,
cosec A = 1/sin A = 1/0 = not defined
sec A = 1/cos A = 1/1 = 1
cot A = 1/tan A = 1/0 = not defined
In the same way, we can find the values of trigonometric ratios for a 90degree angle. Here, angle C is reduced to 0, and the side AB will be nearing side BC such that angle A is almost 90 degrees and AB is almost 0.
Example 1: Evaluate 2 tan^{2} 45° + cos^{2} 30° – sin^{2} 60°.
Solution: Since we know,
tan 45° = 1
cos 30° = √3/2
sin 60° = √3/2
Therefore, putting these values in the given equation:
2(1)2 + (√3/2)2 – (√3/2)2
= 2 + 0
= 2
Example 2: If tan (A + B) =√3 and tan (A – B) =1/√3, 0° < A + B ≤ 90°; A > B, find A and B.
Solution: Given,
tan (A + B) = √3
As we know, tan 60° = √3
Thus, we can write;
⇒ tan (A + B) = tan 60°
⇒(A + B) = 60° …… (i)
Now again given;
tan (A – B) = 1/√3
Since, tan 30° = 1/√3
Thus, we can write;
⇒ tan (A – B) = tan 30°
⇒(A – B) = 30° ….. (ii)
Adding the equation (i) and (ii), we get;
A + B + A – B = 60° + 30°
2A = 90°
A= 45°
Now, put the value of A in eq. (i) to find the value of B;
45° + B = 60°
B = 60° – 45°
B = 15°
Therefore A = 45° and B = 15°
Example 3: Show that :
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution:
(i) tan 48° tan 23° tan 42° tan 67°
We can also write the above given tan functions in terms of cot functions, such as;
tan 48° = tan (90° – 42°) = cot 42°
tan 23° = tan (90° – 67°) = cot 67°
Hence, substituting these values, we get
= cot 42° cot 67° tan 42° tan 67°
= (cot 42° tan 42°) (cot 67° tan 67°)
= 1 × 1 [since cot A.tan A = 1]
= 1
(ii) cos 38° cos 52° – sin 38° sin 52°
We can also write the given cos functions in terms of sin functions.
cos 38° = cos (90° – 52°) = sin 52°
cos 52°= cos (90° – 38°) = sin 38°
Hence, putting these values in the given equation, we get;
sin 52° sin 38° – sin 38° sin 52° = 0
Example 4: If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution: Given,
tan 2A = cot (A – 18°)
As we know by trigonometric identities,
tan 2A = cot (90° – 2A)
Substituting the above equation in the given equation, we get;
⇒ cot (90° – 2A) = cot (A – 18°)
Therefore,
⇒ 90° – 2A = A – 18°
⇒ 108° = 3A
A = 108° / 3
Hence, the value of A = 36°
Example 5: If A, B and C are interior angles of a triangle ABC, then show that sin [(B + C)/2] = cos A/2.
Solution:
As we know, for any given triangle, the sum of all its interior angles is equals to 180°.
Thus,
A + B + C = 180° ….(1)
Now we can write the above equation as;
⇒ B + C = 180° – A
Dividing by 2 on both the sides;
⇒ (B + C)/2 = (180° – A)/2
⇒ (B + C)/2 = 90° – A/2
Now, put sin function on both sides.
⇒ sin (B + C)/2 = sin (90° – A/2)
Since,
sin (90° – A/2) = cos A/2
Therefore,
sin (B + C)/2 = cos A/2
Some Trigonometric Identities are:
(i) sin^{2}θ + cos^{2}θ = 1
(ii) sec^{2}θ  tan^{2}θ =1
(iii) cosec^{2}θ  cot^{2}θ = 1
Example 6: Prove the identities:
(i) √[1 + sinA/1 – sinA] = sec A + tan A
(ii) (1 + tan2A/1 + cot2A) = (1 – tan A/1 – cot A)2 = tan2A
Solution:
(i) Given:√[1 + sinA/1 – sinA] = sec A + tan A
(ii) Given: (1 + tan2A/1 + cot2A) = (1 – tan A/1 – cot A)2 = tan2A
LHS:
= (1+tan²A) / (1+cot²A)
Using the trigonometric identities we know that 1+tan²A = sec²A and 1+cot²A= cosec²A
= sec²A/ cosec²A
On taking the reciprocals we get= sin²A/cos²A
= tan²A
RHS:
=(1tanA)²/(1cotA)²
Substituting the reciprocal value of tan A and cot A we get,
= (1sinA/cosA)²/(1cosA/sinA)²
= [(cosAsinA)/cosA]²/ [(sinAcos)/sinA)²] = [(cosAsinA)²×sin²A] /[cos²A. /(sinAcosA)²] = sin²A/cos²A
= tan2A
The values of LHS and RHS are the same.
Hence proved.
Example 7: If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1.
Solution:
Given,
sin θ + cos θ = √3
Squaring on both sides,
(sin θ + cos θ)^{2 }= (√3)^{2}
sin^{2}θ + cos^{2}θ + 2 sin θ cos θ = 3
Using the identity sin^{2}A + cos^{2}A = 1,
1 + 2 sin θ cos θ = 3
2 sin θ cos θ = 3 – 1
2 sin θ cos θ = 2
sin θ cos θ = 1
sin θ cos θ = sin^{2}θ + cos^{2}θ
⇒ (sin^{2}θ + cos^{2}θ)/(sin θ cos θ) = 1
⇒ [sin^{2}θ/(sin θ cos θ)] + [cos^{2}θ/(sin θ cos θ)] = 1
⇒ (sin θ/cos θ) + (cos θ/sin θ) = 1
⇒ tan θ + cot θ = 1
Hence proved.
Example 8: Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
cot 85° + cos 75°
= cot (90° – 5°) + cos (90° – 15°)
We know that cos(90° – A) = sin A and cot(90° – A) = tan A
= tan 5° + sin 15°
(i) sin (90 – θ) = cosθ
(ii) cos (90 – θ) = sinθ
(iii) tan (90 – θ) = cotθ
(iv) cot (90 – θ) = tanθ
(v) sec (90 – θ) = cosecθ
(vi) cosec (90 – θ) = secθ
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