Definitions [1]
When an equation, involving trigonometrical ratios of an angle A, is true for all values of A, the equation is called a trigonometric identity.
Formulae [3]
\[sineA=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\]
\[cosineA=\frac{\mathrm{Base}}{\text{Hypotenuse}}\]
\[tangentA=\frac{\text{Perpendicular}}{\mathrm{Base}}\]
\[cotangent A = \frac{\text{Base}}{\text{Perpendicular}}\]
\[secantA=\frac{\text{Hypotenuse}}{\mathrm{Base}}\]
\[cosecantA=\frac{\text{Hypotenuse}}{\text{Perpendicular}}\]
\[\sin\mathrm{A}=\frac{1}{\mathrm{cosec~A}}\quad\mathrm{and}\quad\mathrm{cosec~A}=\frac{1}{\sin\mathrm{A}}\]
\[\cos\mathrm{A}=\frac{1}{\sec\mathrm{A}}\quad\mathrm{and}\quad\mathrm{sec}\mathrm{A}=\frac{1}{\cos\mathrm{A}}\]
\[\tan\mathrm{A}=\frac{1}{\cot\mathrm{A}}\quad\mathrm{and}\quad\cot\mathrm{A}=\frac{1}{\tan\mathrm{A}}\]
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sinθ⋅cosecθ = 1
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cosθ⋅secθ = 1
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tanθ⋅cotθ = 1
\[tanA=\frac{\sin A}{\cos A}\]
\[cotA=\frac{\cos A}{\sin A}\]
Theorems and Laws [6]
Prove that `sqrt((1 - sin θ)/(1 + sin θ)) = sec θ - tan θ`.
L.H.S. = `sqrt(((1 - sin θ)(1 - sin θ))/((1 + sin θ)(1 - sin θ)))`
= `sqrt((1 + sin^2θ - 2sinθ)/(1 - sin^2θ)`
= `sqrt((1 + sin^2θ - 2sinθ)/(cos^2θ)`
= `sqrt( 1/cos^2θ + sin^2θ/cos^2θ - (2sin θ)/cos θ xx 1/cosθ`
= `sqrt( sec^2θ + tan^2 θ - 2 tan θ . sec θ)`
= `sqrt((sec θ - tan θ)^2)`
= sec θ – tan θ
= R.H.S.
Hence proved.
L.H.S. = `sqrt((1 - sin θ)/(1 + sin θ))`
= `sqrt(((1 - sin θ)(1 - sin θ))/((1 + sin θ)(1 - sin θ))`
= `sqrt(((1 - sin θ)^2)/(1 - sin^2θ)`
= `sqrt(((1 - sin θ)^2)/(cos^2θ)`
= `(1 - sin θ)/(cos θ)`
= `1/(cos θ) - (sin θ)/(cos θ)`
= sec θ – tan θ
= R.H.S.
Hence Proved.
`1/((1+ sin θ)) + 1/((1 - sin θ)) = 2 sec^2 θ`
LHS =`1/((1+ sin θ)) + 1/((1 - sin θ))`
= `((1 - sin θ) + (1 + sin θ))/((1 + sin θ)(1 - sin θ))`
= `2/(1 - sin^2 θ)`
= `2/(cos^2 θ)`
= 2 sec2 θ
= RHS
Hence Proved.
Prove that:
`tanA/(1 - cotA) + cotA/(1 - tanA) = secA "cosec" A + 1`
L.H.S. = `tanA/(1 - cotA) + cotA/(1 - tanA)`
= `tanA/(1 - 1/tanA) + (1/tanA)/(1 - tanA)`
= `tan^2A/(tanA - 1) + 1/(tanA(1 - tanA))`
= `(tan^3A - 1)/(tanA(1 - tanA))`
= `((tanA - 1)(tan^2A + 1 + tanA))/(tanA(tanA - 1)`
= `(sec^2A + tanA)/tanA`
= `(1/cos^2A)/(sinA/cosA) + 1`
= `1/(sinAcosA) + 1`
= sec A cosec A + 1 = R.H.S.
If 1 + sin2θ = 3 sin θ cos θ, then prove that tan θ = 1 or `1/2`.
Given: 1 + sin2 θ = 3 sin θ cos θ
Dividing L.H.S and R.H.S equations with sin2θ,
We get,
`(1 + sin^2 theta)/(sin^2 theta) = (3 sin theta cos theta)/(sin^2 theta)`
`\implies 1/(sin^2 theta) + 1 = (3 cos theta)/sintheta`
cosec2 θ + 1 = 3 cot θ
Since, cosec2 θ – cot2 θ = 1
`\implies` cosec2 θ = cot2 θ + 1
`\implies` cot2 θ + 1 + 1 = 3 cot θ
`\implies` cot2 θ + 2 = 3 cot θ
`\implies` cot2 θ – 3 cot θ + 2 = 0
Splitting the middle term and then solving the equation,
`\implies` cot2 θ – cot θ – 2 cot θ + 2 = 0
`\implies` cot θ(cot θ – 1) – 2(cot θ + 1) = 0
`\implies` (cot θ – 1)(cot θ – 2) = 0
`\implies` cot θ = 1, 2
Since,
tan θ = `1/cot θ`
tan θ = `1, 1/2`
Hence proved.
Given, 1 + sin2 θ = 3 sin θ cos θ
On dividing by sin2 θ on both sides, we get
`1/(sin^2θ) + 1 = 3 cot θ` ...`[∵ cot θ = cos θ/sin θ]`
⇒ cosec2 θ + 1 = 3 cot θ
⇒ 1 + cot2 θ + 1 = 3 cot θ
⇒ cot2 θ – 3 cot θ + 2 = 0
⇒ cot2 θ – 2 cot θ – cot θ + 2 = 0
⇒ cot θ (cot θ – 2) – 1(cot θ – 2) = 0
⇒ (cot θ – 2) (cot θ – 1) = 0
⇒ cot θ = 1 or 2
tan θ = 1 or `1/2`
Hence proved.
If sin θ + cos θ = `sqrt(3)`, then prove that tan θ + cot θ = 1.
sin θ + cos θ = `sqrt(3)`
Squaring on both sides:
(sin θ + cos θ)2 = `(sqrt(3))^2`
sin2 θ + cos2 θ + 2 sin θ cos θ = 3
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
L.H.S = tan θ + cot θ
= `sin theta/cos theta + cos theta/sin theta`
= `(sin^2 theta + cos^2 theta)/(sin theta cos theta)`
= `1/(sin theta cos theta)`
= `1/1` ...(sin θ cos θ = 1)
= 1 = R.H.S.
⇒ tan θ + cot θ = 1
L.H.S = R.H.S
Prove the following trigonometry identity:
(sin θ + cos θ)(cosec θ – sec θ) = cosec θ ⋅ sec θ – 2 tan θ
LHS = (sin θ + cos θ)(cosec θ – sec θ)
= `(sin θ + cos θ)(1/sin θ - 1/cos θ)`
= `(sin θ + cos θ)((cos θ - sin θ)/(sin θ * cos θ))`
= `(cos^2θ - sin^2θ)/(sinθ * cosθ)`
= `(1 - 2sin^2θ)/(sinθ*cosθ)`
= `1/(sinθ * cosθ) - (2 sin^2θ)/(sinθ * cosθ)`
= `cosec θ · sec θ - (2 sin^2 θ)/(sin θ * cos θ)`
= cosec θ · sec θ – 2 tan θ
= RHS
Hence proved.
Key Points
For an acute angle A in a right-angled triangle:
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Hypotenuse is the side opposite the right angle.
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Perpendicular is the side opposite angle A.
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Base is the side adjacent to angle A.
| Angle | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan | 0 | 1/√3 | 1 | √3 | Not defined |
| cosec | Not defined | 2 | √2 | 2/√3 | 1 |
| sec | 1 | 2/√3 | √2 | 2 | Not defined |
| cot | Not defined | √3 | 1 | 1/√3 | 0 |
sin2 A + cos2 A = 1
1 + tan2 A = sec2 A
1 + cot2 A = cosec2 A
Important Questions [49]
- If θ is an acute angle of a right angled triangle, then which of the following equation is not true?
- If θ is an acute angle and sin θ = cos θ, find the value of tan2 θ + cot2 θ – 2.
- Prove that AAAAAsinA-2sin3A2cos3A-cosA=tanA
- Evaluate 2 sec2 θ + 3 cosec2 θ – 2 sin θ cos θ if θ = 45°.
- If sin θ – cos θ = 0, then find the value of sin4 θ + cos4 θ.
- Prove that θθθθtanθ1-cotθ+cotθ1-tanθ = 1 + sec θ cosec θ
- (3 sin2 30° – 4 cos2 60°) is equal to ______.
- In a right triangle PQR, right angled at Q. If tan P = 3, then evaluate 2 sin P cos P.
- In ΔBC, right angled at C, if tan A = 87, then the value of cot B is ______.
- Evaluate: 5 cosec2 45° – 3 sin2 90° + 5 cos 0°.
- AA1+tan2A1+cot2A is equal to ______.
- If 4 tan θ = 3, evaluate (4sin𝜃−cos𝜃+14sin𝜃+cos𝜃−1)
- Evaluate: (Tan 65°)/(Cot 25°)
- Express (Sin 67° + Cos 75°) in Terms of Trigonometric Ratios of the Angle Between 0° and 45°.
- Prove that (Sin θ + Cosec θ)2 + (Cos θ + Sec θ)2 = 7 + Tan2 θ + Cot2 θ.
- If Sec θ = X + 1/(4"X"), X ≠ 0, Find (Sec θ + Tan θ)
- Evaluate: Tan 65 ∘ Cot 25 ∘
- Prove That: Sqrt(( Secθ - 1)/(Secθ + 1)) + Sqrt((Secθ + 1)/(Secθ - 1)) = 2cosecθ
- Prove That: √ Sec θ − 1 Sec θ + 1 + √ Sec θ + 1 Sec θ − 1 = 2 Cos E C θ
- Prove that: 2(sin^6 θ + cos^6 θ) – 3 (sin^4 θ + cos^4 θ) + 1 = 0.
- A Moving Boat is Observed from the Top of a 150 M High Cliff Moving Away from the Cliff. the Angle of Depression of the Boat Changes from 60° to 45° in 2 Minutes. Find the Speed of the Boat in M/Min.
- There Are Two Poles, One Each on Either Bank of a River Just Opposite to Each Other. One Pole is 60 M High. from the Top of this Pole, the Angle of Depression of the Top And.
- If Sec θ + Tan θ = M, Show that M 2 − 1 M 2 + 1 = Sin θ
- Prove that sqrt((1 – sin θ)/(1 + sin θ)) = sec θ – tan θ.
- Prove that: sinA+cosAsinA-cosA+sinA-cosAsinA+cosA=22sin2A-1
- If sin θ + cos θ = sqrt(3), then prove that tan θ + cot θ = 1.
- If 1 + sin^2θ = 3 sin θ cos θ, then prove that tan θ = 1 or 1/2.
- Show that (cos^2(45^circ + θ) + cos^2(45^circ – θ))/(tan(60^circ + θ) tan(30^circ – θ)) = 1
- If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.
- Θθθcos2θsin2θ-1sin2θ, in simplified form, is ______.
- Proved that 1+secAsecA=sin2A1-cosA.
- If tan θ = xy, then cos θ is equal to ______.
- Sec θ when expressed in term of cot θ, is equal to ______.
- Which of the following is true for all values of θ (0° ≤ θ ≤ 90°)?
- (sec2 θ – 1) (cosec2 θ – 1) is equal to ______.
- Prove that 1+tan2A1+cot2A = sec2 A – 1
- Prove that cotA-cosAcotA+cosA=cos2A(1+sinA)2
- Prove that (sec θ + tan θ) (1 – sin θ) = cos θ
- Prove the identity (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.
- Prove the following trigonometric identities. sec A (1 − sin A) (sec A + tan A) = 1
- Prove the following trigonometric identities. 1+secθsecθ=sin2θ1-cosθ
- 1/((1+ sin θ)) + 1/((1 – sin θ)) = 2 sec^2 θ
- Prove that (Sinθ - Cosθ + 1)/(Sinθ + Cosθ - 1) = 1/(Secθ - Tanθ)
- Cos4 A − sin4 A is equal to ______.
- Find a If Tan 2a = Cot (A-24°).
- Find the value of ( sin2 33° + sin2 57°).
- Prove that :(Sinθ+Cosecθ)2+(Cosθ+ Secθ)2 = 7 + Tan2 θ+Cot2 θ.
- Prove that: (1+cot A - cosecA)(1 + tan A+ secA) =2.
- Prove that Tan 2 a Tan 2 a − 1 + Cos E C 2 a Sec 2 a − Cos E C 2 a = 1 1 − 2 C O 2 a
