Question

prove that cos 11°+ sin11°÷cos11°-sin11°= tan 56°
​

Answer 1

Answer:

as show in photo attached. trick is to identify tan 45 = 1 and applying formulae for Tan(a+b)

Step-by-step explanation:

as shown.

Answer 2

Answer:

Using the facts that sin(45) = cos(45) = 1/(sqrt(2))

(sin(11) + cos(11))= sqrt(2){ sin(11)cos(45) + cos(11)sin(45) } = sqrt(2)(sin(56))

(cos(11) - sin(11)) = sqrt(2){ cos(11)cos(45) - sin(11)sin(45) } = sqrt(2)(cos(56))

Dividing the two identities, (sin(11) + cos(11))/(cos(11) - sin(11)) = tan(56)

OR

cos11°+sin11°cos11°−sin11°

cos11°+sin11°cos11°−sin11° =1+tan11°1−tan11°(dividing both numerator and denominator by cos 11°)

cos11°+sin11°cos11°−sin11° =1+tan11°1−tan11°(dividing both numerator and denominator by cos 11°)=tan45°+tan11°1−tan45°tan11°(∵tan45°=1)

cos11°+sin11°cos11°−sin11° =1+tan11°1−tan11°(dividing both numerator and denominator by cos 11°)=tan45°+tan11°1−tan45°tan11°(∵tan45°=1)=tan(45°+11°)

cos11°+sin11°cos11°−sin11° =1+tan11°1−tan11°(dividing both numerator and denominator by cos 11°)=tan45°+tan11°1−tan45°tan11°(∵tan45°=1)=tan(45°+11°)[∵tan(A+B)=tanA+tanB1−tanAtanB]

cos11°+sin11°cos11°−sin11° =1+tan11°1−tan11°(dividing both numerator and denominator by cos 11°)=tan45°+tan11°1−tan45°tan11°(∵tan45°=1)=tan(45°+11°)[∵tan(A+B)=tanA+tanB1−tanAtanB]=tan56°

please mark me as brainly.......