Question

if tan 15°= 2-√3 then what is the value of 2tan 1095° + cot975° +tan (-195°)​

Answer 1

$\sf { \underline{SOLUTION}}$

GIVEN

tan 15° = 2-√3

TO DETERMINE

The value of

2tan 1095° + cot975° +tan (-195°)

EVALUATION

2 tan 1095° + cot975° + tan (-195°)

= 2 tan 1095° + cot975° - tan 195°

= 2 tan ( 1080° + 15° ) + cot ( 990° - 15° ) -

tan ( 180° + 15°)

= 2 tan ( 12 × 90° + 15° ) + cot ( 11 × 90° - 15° ) - tan ( 2 × 90° + 15°)

= 2 tan 15° - tan 15° - tan 15°

= 2 tan 15° - 2 tan 15°

= 0

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Answer 2

SOLUTION

SOLUTION

SOLUTION

SOLUTION GIVEN

SOLUTION GIVENtan 15° = 2-√3

SOLUTION GIVENtan 15° = 2-√3TO DETERMINE

SOLUTION GIVENtan 15° = 2-√3TO DETERMINEThe value of

SOLUTION GIVENtan 15° = 2-√3TO DETERMINEThe value of2tan 1095° + cot975° +tan (-195°)

SOLUTION GIVENtan 15° = 2-√3TO DETERMINEThe value of2tan 1095° + cot975° +tan (-195°)EVALUATION

SOLUTION GIVENtan 15° = 2-√3TO DETERMINEThe value of2tan 1095° + cot975° +tan (-195°)EVALUATION2 tan 1095° + cot975° + tan (-195°)

SOLUTION GIVENtan 15° = 2-√3TO DETERMINEThe value of2tan 1095° + cot975° +tan (-195°)EVALUATION2 tan 1095° + cot975° + tan (-195°)= 2 tan 1095° + cot975° - tan 195°

SOLUTION GIVENtan 15° = 2-√3TO DETERMINEThe value of2tan 1095° + cot975° +tan (-195°)EVALUATION2 tan 1095° + cot975° + tan (-195°)= 2 tan 1095° + cot975° - tan 195°= 2 tan ( 1080° + 15° ) + cot ( 990° - 15° ) -

SOLUTION GIVENtan 15° = 2-√3TO DETERMINEThe value of2tan 1095° + cot975° +tan (-195°)EVALUATION2 tan 1095° + cot975° + tan (-195°)= 2 tan 1095° + cot975° - tan 195°= 2 tan ( 1080° + 15° ) + cot ( 990° - 15° ) -tan ( 180° + 15°)

SOLUTION GIVENtan 15° = 2-√3TO DETERMINEThe value of2tan 1095° + cot975° +tan (-195°)EVALUATION2 tan 1095° + cot975° + tan (-195°)= 2 tan 1095° + cot975° - tan 195°= 2 tan ( 1080° + 15° ) + cot ( 990° - 15° ) -tan ( 180° + 15°)= 2 tan ( 12 × 90° + 15° ) + cot ( 11 × 90° - 15° ) - tan ( 2 × 90° + 15°)

= 2 tan 15° - tan 15° - tan 15°

= 2 tan 15° - tan 15° - tan 15°= 2 tan 15° - 2 tan 15°

= 2 tan 15° - tan 15° - tan 15°= 2 tan 15° - 2 tan 15°= 0

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