Question

HEYA... ☺☺.
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»TWO QUESTIONS

»SOLVE IT WITH REASON

»10 POINTS

»DONT ANSWER IF U DONT KNOW

Answer 1

Howdy!!

your answer is ----

Given,

(22) A = tan6° tan42° & B = cot66° cot78°

we know that,

tan6° tan42° tan66° tan78° = 1 ..(1)

=> tan6° tan42° = 1/(tan66° tan78°)

=> tan6° tan42° = cot66° cot78° [since,
1/ tan∅ = cot∅]

from (1)

=> A = B

so,option (C) is right .

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23)

64√3sinπ/48cosπ/48cosπ/24cosπ/12cosπ/6

using , 2sinA×cosA= sin2A

= 32√3sin2π/48cosπ/24cosπ/12cosπ/6

=16√3sin2π/24cosπ/12cosπ/6

=16√3sin2π/24cosπ/12cosπ/6

= 8√3sin2π/12cosπ/6

= 4√3sin2π/6

= 4√3sin60°

= 4√3√3/2

=2×3

= 6

so, option (B) is right


hope it help you

Answer 2

Hey !!!

Q .22

A = tan6°tan42° ----------1)

and B = cot66° cot78° ------2)

divide 1st to 2nd equation


A/B = tan6°tan42°/cot66°cot78°

A/B = tan6° tan42°*tan66°×tan78°

•°• tan6° tan42° tan66° tan78° = 1 (you can find it urself , lil complex calculation so just remember it )

hence ,

A/B = 1

A = B (prooved )

since , option (c) is correct Answer ✔

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23 rd no. question ✒

64√3 sinπ/48cosπ/48cosπ/24°cosπ/12cosπ/6

=> 32√3 ×2sinπ/48 cosπ/48 cosπ/24 cosπ/12 cosπ/6

using 2sinA×cosA = sin2A

=> 32√3 sin2π/48 cosπ/24 cosπ/12 cosπ/6

=> 16√3 2sinπ/24 cosπ/24 cosπ/12 cosπ/6

=> 16√3sin2π/24 cosπ/12 cosπ/6

=> 8√3 2sinπ/12 cosπ/12 cosπ/6

=> 8√3 sin2π/12 cosπ/6

=> 4√3 2sinπ/6 cosπ/6

=> 4√3 sin2π/6

=> 4√3 sinπ/3

=> 4√3 sin60°

=> 4√3 √3/2

=> 2×3

=> 6 Answer ✔

Hence option [B ] is correct Answer ✔

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Hope it helps you !!!


@Rajukumar111