Question

in a triangle ABC prove that
b cos C + c cos B = a


don't spam for god's sake and please give proper solution ​

Answer 1

To prove ;-

• b cos C +c cos B = a

─━─━─━─━─━─━─━─━─━─━─━─━─

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

Cosine Law :-

$\tt \:  ⟼ \: cosB \: = \dfrac{ {c}^{2} + {a}^{2} - {b}^{2} }{2ac}$

$\tt \:  ⟼ \: cos \: C \: = \: \dfrac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline\purple{\bold{Solution :- }}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\tt \:  ⟼ \: Consider \: LHS$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\tt \:  ⟼ \: b \: cosC + c \: cosB$

$\tt \:  ⟼ \: = b \: \times \bigg(\dfrac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab} \bigg) + c \times \bigg( \dfrac{ {c}^{2} + {a}^{2} - {b}^{2} }{2ac} \bigg)$

$\tt \:  ⟼ \: = \: \dfrac{ {a}^{2} + {b}^{2} - {c}^{2} }{2a} + \dfrac{ {c}^{2} + {a}^{2} - {b}^{2} }{2a}$

$\tt \:  ⟼ \: \dfrac{ {a}^{2} + {b}^{2} - {c}^{2} + {c}^{2} + {a}^{2} - {b}^{2} }{2a}$

$\tt \:  ⟼ \: = \: \dfrac{ {2a}^{2} }{2a}$

$\tt \:  ⟼ \: = \: a$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large \red{\tt\: ⟼ Explore \: \: more } ✍$

Trigonometry Formulas

sin(−θ) = −sin θ

cos(−θ) = cos θ

tan(−θ) = −tan θ

cosec(−θ) = −cosecθ

sec(−θ) = sec θ

cot(−θ) = −cot θ

Product to Sum Formulas

sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

cos x cos y = 1/2[cos(x–y) + cos(x+y)]

sin x cos y = 1/2[sin(x+y) + sin(x−y)]

cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas

sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Sum or Difference of angles

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

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

sin (A -B) = sin A cos B – cos A sin B

tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]

tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]

cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A

sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

Multiple and Submultiple angles

sin2A = 2sinA cosA = [2tan A /(1+tan²A)]

cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]

tan 2A = (2 tan A)/(1-tan²A)

─━─━─━─━─━─━─━─━─━─━─━─━─