Question

if 7 sin^2 theta + 3 cos^2 theta=4 find the value of sec theta + cosec theta

Answer 1

sec theta. +cosec theta =5/2under root 2

Answer 2

Answer:

$\large{\underline{\boxed{\sf \dfrac{2(1+\sqrt{3}) }{\sqrt{3}}}}}$

Step-by-step explanation:

Given that ;

If 7 sin²θ + 3 cos²θ = 4. Find the value of Secθ + cosecθ.

Here we go;

$7 \sin {}^{2} \theta + 3 \cos {}^{2} \theta = 4$

It can be written as -

$4 \sin^{2} \theta + 3 \sin^{2} \theta+ 3 \cos^{2} \theta = 4 \\ \\ 4 \sin^{2} \theta + 3(\sin^{2} \theta+ \cos^{2} \theta) = 4$

We know that ;

sin²θ + cos²θ = 1, put the value in above.

$4 \sin^{2} \theta + 3(1) = 4 \\ \\4 \sin^{2} \theta = 4 - 3 \\ \\4 \sin^{2} \theta =1 \\ \\ \sin^{2} \theta = \frac{1}{4} \\ \\ \sin \theta = \pm \sqrt{ \frac{1}{4} } \\ \\ \sin \theta = \pm \frac{1}{2}$

Taking the positive value.

$\sin \theta = \dfrac{1}{2} = \dfrac{ \text p}{ \text h}$

Find base ;

$\text b = \sqrt{ \text h {}^{2} - \text p {}^{2} } \\ \\ \text b = \sqrt{(2) {}^{2} - (1) {}^{2} } \\ \\ \text b = \sqrt{4 - 1} \\ \\ \text b = \sqrt{3}$

Now, find the other required t-ratios.

$\sec \theta = \frac{ \text h}{ \text b} = \frac{2}{ \sqrt{3} } \\ \\ \cosec \theta = \frac{ \text h}{ \text p} =2$

Now, we have to find ;

$\sec \theta + \cosec \theta \\ \\ \implies \frac{2}{ \sqrt{3} } + 2 \\ \\ \implies \frac{2 + 2 \sqrt{3} }{ \sqrt{3} } \\ \\ \implies \frac{2(1 + \sqrt{3} )}{ \sqrt{3} }$

Hence, the answer is found.