Question

prove no.2
please do fast

Answer 1

Hey !!!

Q. no. 1

( a)

from Rhs

tan75° - cot75°

tan75° - cot ( 90° - 15 ° )

tan75° - tan15°

sin75°/cos75° - sin15° / cos15°

sin75° × cos15 ° - cos 75°× cos15° / cos75° sin15°

APPLYING formula

sin ( A - B ) = sinA cosA - cosA sinA
and , 2sinA × cosA = sin2A

hence , sin( 75° - 15° ) / cos75° sin15°

multiplying by 2 on both denominator and numerator

=> 2sin 60° / 2 sin15° × cos75°

=> 2sin60° / sin30°

=> 2 sin60° / 1 / 2

=> 4 sin60° Prooved

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(b )

from LHS

cosa + cos ( 120° + a ) + cos ( 120 ° - a )

Applying formula

cosC + cos D = 2 cos ( C + D ) / 2 × sin ( c - D )/2

=> cosa + cos ( 120° + a ) + cos ( 120° - a)

=> cosa + 2 cos { 120° + a + ( 120° - a )/2 } × cos ( 120° + a - 120 + a )/2


=> cosa + 2 cosa 120° × cosA

=> cosa + 2cos ( 90° + 30° ) × cosa

=> cosa - 2 sin30°× cosa

=> cosa - cosa

=> 0 Rhs prooved

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°•° tan ( A + B ) = tanA + tanB / 1 - tanA tanB
APPLYING this formule

c) •°•tan ( 35° + 10 ) = tan45°


tan 30° + tan10° / 1 - tan35°tan10° = 1

tan30° + tan10° = 1 - tan35° tan10°

tan30° + tan10° + tan35° tan10° = 1 Rhs prooved

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( D ) tan ( 8a ) = tan(5a + 3a )

tan ( 5a + 3a ) = tan5a + tan3a/ 1 - tan5a tan3a

tan8a = tan5a + tan3a/ 1 - tan5a tan3a

tan8a - tan8a tan5a tan3a = tan5a + tan3a

tan8a -( tan5a + tan3a) = tan8a tan5a tan3a

tan8a - tan5a - tan3a = tan8a tan5a tan3a

Rhs = LHS prooved


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Hope it will help you !!

@Rajukumar111