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Introduction to Trigonometry

    Exercise 8.3

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10th class > Introduction to Trigonometry > Exercise 8.3

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 = 11+cot2A

sin2A = 11+cot2A ( 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 negative.

Therefore, sin A = 11+cot2A

We know that,

tan A = sin Acos A

However, we have,

cot A = cos Asin A

Therefore, we have,

tan A = 1cot 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−1sec2A

=sec2A−1sec2A.... (2)

tan2A + = sec2A (Trigonometric identity)

tan2A = - (By transposing)

tan A = sec2A−1 ...... (3)

cot A =

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

=1sec2A−1

cosec A = 1sinA

= 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

= 9 (sec2A - tan2A)

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

=

(ii)

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

Answer:

Justify:

We know that using the trigonometric ratios,

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

cot (x) = cos (x)sin(x) = 1tan (x)

sec (x) = 1cos (x)

cosec (x) = 1sin (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 = (a + b)(a - b)]

= sin2θ+cos2θ+2sinθcosθ−1sinθ cosθ

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

= 2 sinθ cosθ sinθ cosθ

=

(iii)

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

Answer:

Justify:

We know that,

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

sec(x) = 1cos(x)

By substituting these values in the given expression we get,

= 1cos A+sin Acos A1−sinA

= 1+sinAcos A1−sinA

= 1−sin2Acos A

= cos^2 Acos A (By using the identity sin2θ + cos2θ = )

=

(iv)

(iv) 1+tan2A1+cot2A

Answer:

Justify:

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

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

= 1tan (x)

By substituting these in the given expression we get,

1+tan2A1+cot2A=1+sin2Acos2A1+cos2Asin2A

=cos2A+sin2Acos2Asin2A+cos2Asin2A

{.reveal(when="blank-12")}= 1cos2A1sin2A [By using the identity: sin^2 A + cos^2 A = ]

= sin^2 Acos^2 A

=