Question

Q24: Cosec θ + Cot θ is equal to
1 point
(a) 1/(Cosec θ - Cot θ)
(b) 1/(Cosec θ + Cot θ)
(c) 1/(Cosec θ )
(d) 1/(Cot θ )
Q25: Cosec θ + Cot θ is equal to
1 point
(a) 1/(Cosec θ - Cot θ)
(b) 1/(Cosec θ + Cot θ)
(c) 1/(Cosec θ )
(d) 1/(Cot θ )
Q26: (a³- b³)= is equal to
1 point
(a) (a+b)(a²+ ab+ b²)
(b) (a-b)(a²+ ab+ b²)
(c) (a+b)(a²- ab+ b²)
(d) (a+b)(a²- ab- b²)
Q27: Cot²A=___________
1 point
(a) - Cosec²A+1
(b) - Cosec²A-1
(c) Cosec²A+1
(d) Cosec²A-1
Q28: If sinθ=x then Cosecθ=________
1 point
(a) 1/x
(b) x²
(c) 1/x²
(d) x
Q29: If Tan θ=12/5 then Secθ=________
1 point
(a) 13
(b) 5
(c) 5/13
(d) 13/5
Q30: If Cos θ=3/5,find the value of Cotθ+ Cosecθ
1 point
(a) 1
(b) 2
(c) 3
(d) 4​

Answer 1

Q.24: option (a) is the right answer of this question,

Q.26: a³-b³=(a-b)(a²+ab+b²)

So, option (a) is the right answer.

Q.27: we already know that, 1+cot²A=cosec²A

or, cot²A=cosec²A-1

so, option (d) is the right answer.

Q.28: we already know that,

sinA=1/cosecA

or, x=1/cosecA [ sinA=x ]

or, cosecA×x=1

or, cosecA= 1/x

So, option (a) is the right answer.

Q.29: tanA=12/5

we already know that,

1+tan²A=sec²A

or, sec²A=1+(12/5)²

or, sec²A=1+144/25

or, sec²A=(25+144)/25

or, sec²A= 169/25

or sec²A= (13/5)²

or, secA=13/5

So, option (d) is the right answer.

Q.30: cosA=3/5

so, secA=5/3

we already know that,

1+tan²A=sec²A

or, 1+tan²A=(5/3)²

or, tan²A= 25/9-1

or, tan²A= (25-9)/9

or, tan²A= 16/9

or, tan²A=(4/3)²

or, tanA=4/3

So, cotA=3/4 ----------- (I)

then, we already know that ,

1+cot²A=cosec²A

or, cosec²A=1+(3/4)²

or, cosec²A=1+9/16

or, cosec²A=(16+9)/16

or,cosec²A=25/16

or, cosec²A=(5/4)²

or, cosecA= 5/4 ------------ (II)

So, cotA+cosecA=3/4+5/4

or, cotA+cosecA=(3+5)/4

or, cotA+cosecA=8/4

or, cotA+cosecA=2

So, option (b) is the right answer of this question.

I hope , its helpful for you.