Question

Tan theta=20/21then find the value of 1-sin theta+cos theta/1+sin theta+cos theta

Answer 1

Given

⇒Tanθ = 20/21

To Find

⇒(1-Sinθ + Cosθ)/(1+Sinθ+Cosθ)

First of all We have to find Sinθ and Cosθ

So take

⇒Tanθ = 20/21 = Perpendicular(p)/Base(b)

We get

⇒Perpendicular = 20 , Base(b) = 21 and Hypotenuse(h) = h

Using Pythagoras theorem

⇒h² = p² + b²

⇒h² = (20)² + (21)²

⇒h² = 400 + 441

⇒h² = 841

⇒h = √(841)

⇒h = 29

We get

⇒Perpendicular = 20 , Base(b) = 21 and Hypotenuse(h) = 29

Then

⇒Sinθ = P/h and Cosθ = b/h

⇒Sinθ = 20/29 and Cosθ  = 21/29

Now Put the value on

⇒(1-Sinθ + Cosθ)/(1+Sinθ+Cosθ)

⇒(1-20/29 + 21/29)/(1+20/29 + 21/29)

⇒{(29-20+21)/29}/{29+20+21)/29}

⇒{(50 - 20)/29}/{(50+20)/29}

⇒(30/29)/(70/29)

⇒30/29 ×29/70

⇒30/70

⇒3/7

Answer = 3/7

Answer 2

Step-by-step explanation:

ANSWER ✍️

Given

⇒Tanθ = 20/21

To Find

⇒(1-Sinθ + Cosθ)/(1+Sinθ+Cosθ)

First of all We have to find Sinθ and Cosθ

So take

⇒Tanθ = 20/21 = Perpendicular(p)/Base(b)

We get

⇒Perpendicular = 20 , Base(b) = 21 and Hypotenuse(h) = h

Using Pythagoras theorem

⇒h² = p² + b²

⇒h² = (20)² + (21)²

⇒h² = 400 + 441

⇒h² = 841

⇒h = √(841)

⇒h = 29

We get

⇒Perpendicular = 20 , Base(b) = 21 and Hypotenuse(h) = 29

Then

⇒Sinθ = P/h and Cosθ = b/h

⇒Sinθ = 20/29 and Cosθ  = 21/29

Now Put the value on

⇒(1-Sinθ + Cosθ)/(1+Sinθ+Cosθ)

⇒(1-20/29 + 21/29)/(1+20/29 + 21/29)

⇒{(29-20+21)/29}/{29+20+21)/29}

⇒{(50 - 20)/29}/{(50+20)/29}

⇒(30/29)/(70/29)

⇒30/29 ×29/70

⇒30/70

⇒3/7

Answer = 3/7

hope this helps you!!

thank you ⭐