Question

1) If sec tan ook then proove that
sino-k²-1
k ²+1​

Answer 1

Given data :

if sec θ + tan θ = k then prove that, sin θ = (k² - 1)/(k² + 1)

Proof :

⟹ sin θ = (k² - 1)/(k² + 1)

Here we know, sec θ + tan θ = k

Hence, now

⟹ sin θ = {(sec θ + tan θ)² - 1}/{(sec θ + tan θ)² + 1}

⟹ sin θ = {sec² θ + tan² θ + 2 * sec θ * tan θ - 1}/{sec² θ + tan² θ + 2 * sec θ * tan θ + 1}

⟹ sin θ = {(sec² θ - 1 ) + tan² θ + 2 * sec θ * tan θ }/{sec² θ + (tan² θ + 1) + 2 * sec θ * tan θ }

Here, we know sec² θ + tan² θ = 1

∴ sec² θ - 1 = tan² θ and tan² θ + 1 = sec² θ

⟹ sin θ = { tan² θ + tan² θ + 2 * sec θ * tan θ }/{sec² θ + sec² θ + 2 * sec θ * tan θ }

⟹ sin θ = {2 * tan² θ + 2 * sec θ * tan θ }/ {2 * sec² θ + 2 * sec θ * tan θ }

Now,

⟹ sin θ = {2 * tan θ (sec θ + tan θ) }/ {2 *sec θ (sec θ + tan θ) }

[Here, sec θ + tan θ cancel from numerator and denominator]

⟹ sin θ = {2 * tan θ}/{2 * sec θ}

[Here, 2 get cancel from numerator and denominator]

⟹ sin θ = tan θ/sec θ

Here, we know, 1/sec θ = cos θ

⟹ sin θ = tan θ/sec θ

⟹ sin θ = tan θ * cos θ

Here, we know, tan θ = sin θ/cos θ

⟹ sin θ = sin θ/cos θ * cos θ

[Here, cos θ cancelled]

⟹ sin θ = sin θ

Answer : Therefore, if sec θ + tan θ = k then, sin θ = (k² - 1)/(k² + 1)

Answer 2

Answer:

Sin (x) is an odd function because sin(-x) = -sin(x). It's graph is symmetric to the origin.

Step-by-step explanation:

Sin (x) is an odd function because sin(-x) = -sin(x). It's graph is symmetric to the origin.