Question

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Answer 1

Answer and explanation:

27.

Multiply numerator and denominator with 1+sinA:

$\\\tt \sqrt{ \dfrac{1 + \sin A }{1 - \sin A } \times \dfrac{1 + \sin A }{1 + \sin A } } \\ \\\tt = \sqrt{ \dfrac{(1 + \sin A)^{2} }{1 - \sin ^{2} A } } \\ \\\tt = \dfrac{1 + \sin A }{ \cos A } \\\tt = \sec \: A + \tan \: A$

28.

$\\\tt P(favourable \: event)= \dfrac{no \: of \: favourable \: outcome}{no \: of \: all \: possible \: outcomes }$

i)

Favourable outcome ={2,3,5}

$\\\tt P(event)= \dfrac{3}{6 } = 0.5$

ii) Favourable outcomes ={3,4,5}

$\\\tt P(event)= \dfrac{3}{6 } = 0.5$

iii) Favourable outcome ={1,3,5}

$\\\tt P(event)= \dfrac{3}{6 } = 0.5$

iv) Favourable outcome ={3,6}

$\\\tt P(event)= \dfrac{2}{6 } = \dfrac{1}{3}$

Answer 2

Step-by-step explanation:

Solutions:-

27)

LHS:-

√[(1+SinA)/(1-SinA)]

Multiplying the numerator and the denominator with (1+SinA) then

=> √[(1+SinA)(1+SinA)/(1-SinA)(1+SinA)]

=> √[(1+SinA)^2/(1-SinA)(1+SinA)]

Denominator is in the form of (a+b)(a-b)

Where a = 1 and b= SinA

(a+b)(a-b) = a^2-b^2

=> √[(1+SinA)^2/(1-Sin^2A)]

We know that

Sin^2A+Cos^2A = 1

=> √[(1+SinA)^2/(Cos^2A)]

=> √[(1+SinA)/CosA]^2

=> (1+SinA)/CosA

=> (1/CosA) +(SinA/CosA)

=> Sec A+ Tan A

=> RHS

LHS= RHS

Hence, Proved.

28)

Given that

A die is thrown once

We know that

A die is thrown n times then the total possible outcomes = 6^n

If a die is thrown once then the total possible outcomes = 6^1 = 6

They are 1,2,3,4,5,6

Sample space ={1,2,3,4,5,6}

i) Probability of getting a prime number:-

The prime numbers =2,3,5

Number of favourable outcomes = 3

We Know that

Probability of an event P (E) =

Number of favourable outcomes/Total number of all possible outcomes

Probability of getting a prime number

= 3/6

= 1/2

Probability of getting a prime number = 1/2

ii) Probability of getting a number lying between 2 and 6:-

The numbers between 2 and 6 = 3,4,5

Number of favourable outcomes = 3

We Know that

Probability of an event P (E) =

Number of favourable outcomes/Total number of all possible outcomes

Probability of getting a number lying between 2 and 6

= 3/6

= 1/2

Probability of getting a number lying between 2 and 6 = 1/2

iii) Probability of getting an odd number:-

The odd numbers = 1,3,5

Number of favourable outcomes = 3

We Know that

Probability of an event P (E) =

Number of favourable outcomes/Total number of all possible outcomes

Probability of getting an odd number

= 3/6

= 1/2

Probability of getting an odd number = 1/2

iv) Probability of getting multiple of 3:-

The multiples of 3 = 3 ,6

Number of favourable outcomes = 2

We Know that

Probability of an event P (E) =

Number of favourable outcomes/Total number of all possible outcomes

Probability of getting a multiple of 3

= 2/6

= 1/3

Probability of getting a multiple of 3 = 1/3

Used formulae:-

1.A die is thrown n times then the total possible outcomes = 6^n

2.Probability of an event P (E) =

Number of favourable outcomes/Total number of all possible outcomes