Question

${ \rm{ \huge {Prove \: That}}}$
${ \rm{ \large{cot \: B + cos \: B = sec \: A \: cos \: B(1 + B)}}}$

${ \rm{ \large{Don't \: spam}}}$
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Answer 1

Solution :

= cot B + cos B = LHS

= cot B + cos B = cos B/sin B

= cos B [1/sin B + 1]

= Cos B [1 + sin B/sin B]

= cos B (1 + sin B)/sin(90° - A)

= cos B(1 + sin B) /cos A

= sin A cos B (1 + sin B)

= RHS

Addition information :

tanø = sinø/cosø

secø = 1/cosø

cotø = 1/tanø = sinø/cosø

1 - tan(ø/2)/1 - tan(ø/2) = ±√1 - sinø/1 + sinø

tan ø/2 = ±√1 - cosø/1 + cosø

Answer 2

To Prove :-

• cot B + cos B = sec A cos B (1 + sin B)

$\large{\boxed{\bf{Solution}}}$

★ Since, A and B are complementary angles and A + B = 90°,

L.H.S = cot B + cos B

➪ cot (90° - A) + cos (90° - A)

➪ tan A + sin A

➪ (sin A/cos A) + sin A

➪ (sin A + sin A cos A) / cos A

➪ sin A (1 + cos A) / cos A

➪ sec A sin A (1 + cos A)

➪ sec A sin (90° - B) [1 + cos (90° - B)]

➪ sec A cos B (1 + sin B) = R.H.S

More To Know :-

• tan θ = sin θ/ cosθ

• sec θ = 1/cos θ

• cot θ = 1/ tan θ

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