Math, asked by akumar9038, 1 year ago

without using tables , evaluate 4tan60° sec30° + (sin 31°sec59° + cot59°cot 31°/8sin^2 30° - tan^2 45°)

Answers

Answered by thameshwarp9oqwi
26

Answer:

evaluate 4tan60° sec30° + (sin 31°sec59° + cot59°cot 31°/8sin^2 30° - tan^2 45°)

==>  4tan60° sec30° + (sin 31°sec59° + cot59°cot 31°/8sin^2 30° - tan^2 45°)

==>  4tan60° sec30° + (sin 31°sec(90-31)° + cot59°cot(90-59)°/8sin^2 30° - tan^2 45°)

4tan60° sec30° + (sin 31°cosec31)° + cot59°tan59 /8sin^2 30° - tan^2 45°)

==>  4tan60° sec30° +( sin31*1/sin31 + cot59*1/cot59  /8sin^2 30° - tan^2 45°)

==> 4tan60° sec30° + (1 + 1 /8sin^2 30° - tan^2 45°)

==> 4×√3×2/√3 + [2 / 8(1/2)² - (1)² ]

==> 4×2 ×2  / 8×1/4- 1

==> 16 / 2-1

==> 16/1

==> 16


anupmaagarwal78: This answer is wrong
Answered by 7405413342
12

Answer:

Given Without using tables , evaluate 4tan60° sec30° + sin 31°sec59° + cot59°cot 31°/8sin^2 30° - tan^2 45°

4 tan 60 sec 30 + sin 31 sec 59 + cot 59 cot 31 / 8 sin²30 - tan²45

4 tan 60 sec 30 + sin 31 sec(90 - 31) + cot 59 cot(90 - 59) / 8 sin²30 - tan²45

4 tan 60 sec 30 + sin 31 cosec 31 + cot 59 tan 59 / 8 sin²30 - tan²45

4 tan 60 sec 30 + sin 31 x 1 / sin 31 + cot 59 x 1 /cot 59 / 8 sin²30 - tan²45

4 x √3 x 2 /√3 + 1 + 1/8 x 1/4 - 1)

4 x 2 + 1 + 1

8 + 2 = 10

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