Exercise 8.3

1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A

Solution:

Consider a ΔABC with ∠B = °

Using the Trigonometric Identity,

cosec2 A = + cot2 A (By taking reciprocal both the sides)

1cosec2A =

sin2A = ( As 1cosec2A = sin2A)

Therefore, sin A = ± 11+cot2A

For any sine value with respect to an acute angle in a triangle, the sine value will never be .

Therefore, sin A = 11+cot2A

We know that,

tan A =

However, we have,

cot A =

Therefore, we have,

tan A =

Also, sec2A= + tan2A (Trigonometric Identity)

= + 1cot2A

= cot2A+1cot2A

sec A = cot2A+1cotA

2. Write all the other trigonometric ratios of ∠A in terms of sec A

Solution:

sin2A + cos2A =

cosec2A = + cot2A

sec2A = + tan2A

We know that,

cos A = ..... (1)

Also, sin2A + cos2A = (trigonometric identity)

sin2A = - cos2A (By transposing)

Using value of cos A from Equation (1) and simplifying further

sin A = 1−1secA2

=sec2A−1secA.... (2)

tan2A + = sec2A (Trigonometric identity)

tan2A = - (By transposing)

tan A = ...... (3)

cot A =

= 1secAsec2A−1secA .....(By substituting the values from Equations (1) and (2))

cosec A =

= secAsec2A−1 (By substituting from Equation (2) and simplifying)

3. Answer weather it is true or false. Justify your choice

(i)

(i) 9 sec2A - 9 tan2A =

sin2A + cos2A =

cosec2A = + cot2A

sec2A = + tan2A

Answer:

Justify:

(i) 9sec2A - 9 tan2A

= (sec2A - tan2A)

= 9 × [By using the identity, 1 + sec2A = tan2A]

(ii)

(ii) (1 + tanθ + secθ) (1 + cotθ - cosecθ) =

Answer: (1 + tanθ + secθ) (1 + cotθ - cosecθ) =

Justify:

We know that using the trigonometric ratios,

tan (x) =

cot (x) = cos(x)sin(x) =

sec (x) =

cosec (x) =

By substituting the above function in equation (1),

= 1+sinθcosθ+1cosθ1+cosθsinθ−1sinθ

=cosθ+sinθ+1cosθ sinθ+cosθ−1sinθ (By taking LCM and multiplying)

= sinθ+cosθ2−−12sinθ cosθ (Using a2 - b2 = )

= sin2θ+cos2θ+2sinθcosθ−1sinθ cosθ

= 1+2sinθcosθ−1sinθ cosθ (Using identity sin2θ + cos2θ = )

(iii)

(iii) (sec A + tan A) (1 - sin A)=

Answer:

Justify:

We know that,

tan(x) = sin(x)cos(x)

sec(x) =

By substituting these values in the given expression we get,

= 1cos A+sin Acos A1−sinA

= 1+sinAcos A1−sinA

= cos2AcosA (By using the identity sin2θ + cos2θ = )

(iv)

(iv) 1+tan2A1+cot2A

Answer:

Justify:

tan (x) = sin (x)cos (x)

cot (x) =

cot (x) =

By substituting these in the given expression we get,

1+tan2A1+cot2A=1+sin2Acos2A1+cos2Asin2A

=cos2A+sin2Acos2Asin2A+cos2Asin2A

= 1cos2A1sin2A [By using the identity: sin2A + cos2A = ]

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i)

(i) cosecθ−cotθ2 = 1−cosθ1+cosθ

cosecθ−cotθ2 = cosecθ2 − 2cosecθcotθ + cotθ2

Substituting cosecθ = and cotθ =

cosecθ−cotθ2 = 1sinθ−cosθsinθ2

= 1−cosθ2sin2θ

We can re-write: sin2θ =

= 1−cosθ21−cos2θ

The numerator is 1−cosθ2 =

= 1−2cosθ+cos2θ1−cos2θ

1−cos2θ can be factored as

= 1−2cosθ+cos2θ1+cosθ1−cosθ

Cancel (1 - cosθ) from numerator and denominator

(ii)

(ii) cosA1+sinA + 1+sinAcosA = 2 secA

LHS =

We know: sin2A + =

Therefore: sin2A = 1−cos2A

= cos2A+1+2sinA+1−cos2AcosA1+sinA

= 21+sinAcosA1+sinA

Recall that: secA =

(iii)

(iii) tanθ1−cotθ + cotθ1−tanθ = 1 + secθ cosecθ

[Hint : Write the expression in terms of sinθ and cosθ]

First, recall these identities:

tanθ =

cotθ =

secθ =

cosecθ =

tanθ1−cotθ = sinθcosθ1−cosθsinθ

= cos2θsinθcosθ−sinθ

cotθ1−tanθ =

= cosθsinθcosθcosθ−sinθ

Now, we have:

sin2θcosθsinθ−cosθ + cos2θsinθcosθ−sinθ

= sin2θcosθsinθ−cosθ - cos2θsinθsinθ−cosθ (getting a common denominator)

= sinθ−cosθsin2θ+sinθcosθ+cos2θsinθ·cosθsinθ−cosθ

= sin2θsinθ·cosθ + sinθcosθsinθ·cosθ + cos2θsinθ·cosθ

= sinθcosθ + 1 + cosθsinθ

= + 1 +

Recall that: tanθ + cotθ =

= secθ·cosecθ + 1

= 1 + secθ·cosecθ

(iv)

(iv) 1+secAsecA = sin2A1−cosA

[Hint : Simplify LHS and RHS separately]

LHS 1+secAsecA =

= cosA + 1

RHS: sin2A1−cosA =

= +

(v)

(v) cosA−sinA+1cosA+sinA−1 = cosec A + cot A, using the identity cosec2A=1+cot2A.

Let's start with the LHS: cosA−sinA+1cosA+sinA−1

Let's multiply both numerator & denominator by cosA+sinA+1cosA+sinA+1

= cosA−sinA+1cosA+sinA+1cosA+sinA−1cosA+sinA+1

(cosA - sinA + 1)(cosA + sinA + 1)

= cos2A + + cosA - sinA·cosA - sin2A - + cosA + sinA +

= cos2A - sin2A +

(cosA + sinA - 1)(cosA + sinA + 1) = cos2A + sin2A - 1

Using identity sin2A + cos2A = 1

Denominator becomes: cos2A+sin2A−1 =

For numerator: cos2A - sin2A + 2cosA = 2cos2A - 1 + 2cosA (using same identity)

= 2cos2A2sinA·cosA + 2cosA2sinA·cosA - 12sinA·cosA

= cos2AsinA·cosA + cosAsinA·cosA - 12sinA·cosA

= +

= cosAsinA + 1sinA

= +

= cosecA + cotA

(vi)

(vi) 1+sinA1−sinA = secA + tanA

Squaring both sides:

1+sinA1−sinA = secA+tanA2

RHS: sec2A + 2secA·tanA + tan2A

= + 2secA·tanA + tan2A

= 1 + + 2secA·tanA

LHS: 1+sinA1−sinA = 1+sinAcosA1−sinAcosA

= secA+tanAsecA+tanAsec2A−tan2A

Using identity = 1:

= sec2A + + tan2A

= 1 + 2tan2A + 2secA·tanA

(vii)

(vii) sinθ−2sin3θ2cos3θ−cosθ = tanθ

sinθ−2sin3θ2cos3θ−cosθ =

Recall the identity: sin2θ + cos2θ = 1

Therefore: sin2θ=1−cos2θ

sinθ1−21−cos2θcosθ2cos2θ−1

= sinθ1−2+2cos2θcosθ2cos2θ−1

(viii)

(viii) sinA+cosecA2 + cosA+secA2 = 7 + tan2A + cot2A

sinA+cosecA2 = sin2A + 2sinA·cosecA + cosec2A = sin2A + +

cosA+secA2 = cos2A + 2cosA·secA + sec2A = cos2A + +

sec2A = 1cos2A and cosA·secA =

sin2A+2+1sin2A + cos2A+2+1cos2A = sin2A + cos2A + + 1sin2A + 1cos2A

Using identities:

sin2A+cos2A =

1sin2A =

1cos2A =

sin2A+2+1sin2A + cos2A+2+1cos2A = 1 + 4 + cosec2A + sec2A

tan2A+1 =

cot2A+1 =

1 + 4 + cosec2A + sec2A = 1 + 4 + cot2A+1 + tan2A+1

= + tan2A + cot2A

(ix)

(ix) (cosec A – sin A)(sec A – cos A) = 1tanA+cotA

(Hint : Simplify LHS and RHS separately)

LHS: (cosec A - sin A)(sec A - cos A)

= 1sinA−sinA1cosA−cosA

= 1−sin2AsinA × 1−cos2AcosA

Use Pythagorean identity sin2A+cos2A=1:

1−sin2AsinA × 1−cos2AcosA = ×

= sinA×cosA

RHS: 1tanA+cotA = 1sinAcosA+cosAsinA = = = sin A × cos A

(x)

(x) 1+tan2A1+cot2A = 1−tanA1−cotA2 = tan2A

Using 1+tan2A = sec2A and 1+cot2A = cosec2A

1+tan2A1+cot2A = sec2Acosec2A = 1cos2A1sin2A = =

1−tanA1−cotA2 = 1−tanA1−1tanA2 = × tanA2 =

Chapter 8: Introduction to Trigonometry

Glossary

Exercise 8.3

(i)

(ii)

(iii)

(iv)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

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Chapter 8: Introduction to Trigonometry

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Glossary

Exercise 8.3

(i)

(ii)

(iii)

(iv)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)