Question

If 3 tanA= 4, then prove that underoot 1-sinA divided by 1+cosA = 1 divided 2 root2

Answer 1

Answer:

3tanA=4

or, tanA=4/3=p/b

Using Pythagorus theorem, h²=p²+b² we get,

h²=4²+3²

or, h²=16+9

or, h²=25

or, h=5 (neglecting the negative sign)

∴, secA=h/b=5/3 and cosecA=h/p=5/4

i) √secA-cosecA/√secA+cosecA

=√(5/3-5/4)/(5/3+5/4)

=√(5/12)/√(35/12)

=√(5/12×12/35)

=√1/7

ii) 1-sinA/1+cosA

=(1-4/5)/(1+3/5)

={(5-4)/5}/{(5+3)/5}

=(1/5)/(8/5)

=1/5×5/8

=1/8

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

We are given that

3tanA = 4

tanA = 4/3

To prove

√(1–sinA)/√(1+cosA) = 1/2√2

Proof

tanA = opposite/adjacent

tanA = sinA/cosA

By pythagorean theorem we can find the hypotenuse

(Height)^2+(Base)^2=(Hypotenuse)^2

(3)^2+(4)^2 = (Hypotenuse)^2

9+16 = (Hypotenuse)^2

25 = (Hypotenuse)^2

Hypotenuse = √25

Hypotenuse = 5

sinA = 4/5

cosA = 3/5

LHS RHS

√(1–sinA)/√(1+cosA) = 1/2√2

LHS :

√(1–4/5)/√(1+3/5)

=> √{(5–4)/5}/√{(5+3)/5}

=> √(1/5)/√(8/5)

=> √1/5×5/8

=> √1/8

=> 1/2√2. (Because 8 = √4×√2)

Hence proved

Hope this helps you