Question

if 4sintheta =3,then show that sec theta and cot theta​

Answer 1

Answer:

Step-by-step explanation:

4sinθ=3

or, sinθ=3/4

∴, cosecθ=1/sinθ=4/3

∵, cosec²θ-cot²θ=1

or, cot²θ=cosec²θ-1

or, cot²θ=(4/3)²-1

or, cot²θ=16/9-1

or, cot²θ=7/9

or, cotθ=√7/3

Again, tanθ=1/cotθ=3/√7

∵, sec²θ-tan²θ=1

or, sec²θ=1+tan²θ

or, sec²θ=1+9/7

or, sec²θ=16/7

or, secθ=4/√7

∴, cosθ=1/secθ=√7/4

√(cosec²θ-cot²θ)/(sec²θ-1) + 2cotθ=√7/x + cosθ

or, √[1/{(4/√7)²-1}]+2×√7/3=√7/x+√7/4

or, √{1/(16/7-1)}+2√7/3=√7/x+√7/4

or, √1/(9/7)+2√7/3=√7/x+√7/4

or, √7/3+2√7/3=√7/x+√7/4

or, 1/3+2/3=1/x+1/4

or, -1/x=1/4-1/3-2/3

or, -1/x=1/4-(1/3+2/3)

or, -1/x=1/4-1

or, -1/x=-3/4

or, x=4/3 Ans.

Answer 2

Answer:

4 sin theta = 3

sin theta = 3/4

opp/hypo = 3/4

so opp will be 3k and hypo will be 4k

sec theta =hypo /adjacent

» 4k / √{(4k)²-(3k)²}

( as hypo ² = oppo²+adje² formula)

(adj= √ hypo²-oppo²)

» 4k/ √( 16k² - 9k²)

» 4k/ √7k² ( taking out k² from root)

» 4k/ √7 k

» 4/√7

cot theta = adj/opp

» √7 k/3k

» √7/3

this may help you ...