Question

cos35 - sin35/cos35+sin35=tan10​

Answer 1

$: \implies \tt \: To prove \: \dfrac{cos \: 35 \degree \: - \: sin \: 35 \degree}{cos \: 35 \degree \: + \: sin \: 35 \degree} \: = \: tan \: 10 \degree$

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

We know,

$: \implies \tt \:tan(x - y) = \dfrac{tanx \: - \: tany}{1 + \: tanx \: tany}$

● Consider RHS :-

$: \implies \tt \:tan \: 10 \degree$

$: \implies \tt \:tan(45 \degree \: - 35 \degree)$

$: \implies \tt \:\dfrac{tan \: 45 \degree - tan35 \degree}{1 + tan45 \degree \: tan35 \degree}$

$: \implies \tt \:\dfrac{1 - tan35 \degree}{1 + tan35 \degree}$

$: \implies \tt \:\dfrac{1 - \dfrac{sin35 \degree}{cos35 \degree} }{1 + \dfrac{sin35 \degree}{cos35 \degree} }$

● On taking LCM, we get

$: \implies \tt \:\dfrac{cos \: 35 \degree - \: sin \: 35 \degree}{cos \: 35 \degree + \: sin \: 35 \degree}$

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

Additional Information

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)