Question

Exact value of sec 10° - tan 10º – tan 40° is equal to​

Answer 1

Answer:

sec 10 = 1/cos 10

tan 10 = SIN 10/cos 10

sec 10 - tan 10 = 1-sin 10/cos 10

tan 40 = sin 40/ cos 40

1-sin 10/ cos 10 - sin 40 / cos 40

Answer 2

sec10° - tan10° - tan40° = 0

Concept to be used:

secθ = $\dfrac{1}{cos\theta}$

tanθ = $\dfrac{sin\theta}{cos\theta}$

sin(A + B) = sinA cosB + cosA sinB

sin(90° - θ) = cosθ

cos(90° - θ) = sinθ

Step-by-step explanation:

Now, sec10° - tan10° - tan40°

= $\dfrac{1}{cos10^{o}}-\dfrac{sin10^{o}}{cos10^{o}}-\dfrac{sin40^{o}}{cos40^{o}}$

= $\dfrac{cos40^{o}-sin10^{o}cos40^{o}-sin40^{o}cos10^{0}}{cos10^{o}cos40^{o}}$

= $\dfrac{cos40^{o}-(sin10^{o}cos40^{o}+sin40^{o}cos10^{0})}{cos10^{o}cos40^{o}}$

= $\dfrac{cos40^{o}-sin(10^{o}+40^{o})}{cos10^{o}cos40^{o}}$

= $\dfrac{cos40^{o}-sin50^{o}}{cos10^{o}cos40^{o}}$

= $\dfrac{cos40^{o}-sin(90^{o}-40^{o})}{cos10^{o}cos40^{o}}$

= $\dfrac{cos40^{o}-cos40^{o}}{cos10^{o}cos40^{o}}$

= $\dfrac{0}{cos10^{o}cos40^{o}}$

= 0

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