Question

plzzz help class 9 optional math

If tan theta =p show that sec theta*cosec theta =p+1/p​

Answer 1

Answer:

$sec(\theta) \times cosec(\theta) = \dfrac{tan^2 (\theta)+ 1}{tan (\theta)} = tan (\theta)+ \dfrac{1}{tan (\theta)} = p + \dfrac{1}{p}$

Step-by-step explanation:

The given trigonometric relations are

tan(θ) = p

sec(θ)×cosec(θ) = p + 1/p

We note that, when tan(θ) = p, we have;

p + 1/p = tan(θ) + 1/(tan(θ)) = (tan²(θ) + 1)/tan(θ)

By trigonometric ratios, we have;

tan²(θ) + 1 = sec²(θ) =1/cos²(θ) which gives;

(tan²(θ) + 1)/tan(θ) = 1/cos²(θ) × 1/tan(θ)  = cos(θ)/sin(θ)×1/cos²(θ)  

$\dfrac{1}{cos^2(\theta)} \times \dfrac{cos (\theta)}{sin( \theta)} = \dfrac{1}{cos(\theta)} \times \dfrac{1}{sin( \theta)} = sec(\theta) \times cosec(\theta)$

Therefore;

$If \ tan (\theta) = p \ then \ sec(\theta) \times cosec(\theta) = p + \dfrac{1}{p}$