Question

xsin theta = 1 and r cos= $\sqrt{3$ find $\sqrt{3$ tan theta+ 1

Answer 1

Step-by-step explanation:

We have , ∫π/30tanθ2ksecθ−−−−−−√dθ=1−12–√,(k>0)

Let I=∫π/30tanθ2ksecθ−−−−−−√dθ=12k−−√∫π/30tanθsecθ−−−−√dθ

=12k−−√∫π/30(sinθ)(cos)θ1cosθ−−−−√dθ=12k−−√∫π/30sinθcosθ−−−−√dθ

Let cosθ=t⇒−sinθdθ=dt⇒sinθdθ=−dt

for lower limit , θ=0⇒t=cos0=1

for upper limit , θ=π3⇒t=cosπ3=12

⇒I=12k−−√∫1/21dtt√=−12k−−√∫1/21t−12dt

=−12k−−√(t12+1−12+1)12=−12k−−√[2t√]121

=−22k−−√[12−−√−1–√]1=22k−−√−(1−12–√) ∵I=1−12–√ (given)

∴22k−−√(1−12–√)=1−12–√⇒22k−−√=1

⇒2=2k−−√⇒2k=4⇒k=2

Answer 2

Answer:

rsinθ=1

rcosθ=3–√

⇒ rsinθ rcosθ=13–√

⇒tanθ=13–√

⇒3–√tanθ=1(Add 1 both sides)

⇒3–√tanθ+1=1+1

⇒3–√tanθ+1=2