Question

$\mathbb{ \tt{ \underline { \underline{CBSE \: CLASS - X}}}}$
$\sf{ \underline{Trigonometry}}$

If sec θ + tan θ = p, find cosec θ.

Explainatory answer needed.

Answer 1

Solution :

Given that,

secθ + tanθ = p ...(i)

By identity property of Trigonometry, we know that

sec²θ - tan²θ = 1

⇒ (secθ + tanθ) (secθ - tanθ) = 1

⇒ p (secθ - tanθ) = 1, by (i)

⇒ secθ - tanθ = $\frac{1}{p}$ ...(ii)

Now, adding (i) and (ii) no. equations, we get

2 secθ = p + $\frac{1}{p}$

⇒ secθ = $\frac{p^{2}+1}{2p}$

⇒ cosθ = $\frac{2p}{p^{2}+1}$

Now, sin²θ = 1 - cos²θ

= 1 - $(\frac{2p}{p^{2}+1})^{2}$

= 1 - $\frac{4p^{2}}{p^{4}+2p^{2}+1}$

= $\frac{p^{4}+2p^{2}+1-4p^{2}}{p^{4}+2p^{2}+1}$

= $\frac{p^{4}-2p^{2}+1}{p^{4}+2p^{2}+1}$

= $\frac{(p^{2}-1)^{2}}{(p^{2}+1)^{2}}$

= $(\frac{p^{2}-1}{p^{2}+1})^{2}$

⇒ sinθ = $\frac{p^{2}-1}{p^{2}+1}$

⇒ cosecθ = $\frac{p^{2}+1}{p^{2}-1}$

Useful identities :

1. sinθ * cosecθ = 1

2. sin²θ + cos²θ = 1

3. cosθ * secθ = 1

4. a² - b² = (a + b) (a - b)

#MarkAsBrainliest

Answer 2

Solution :

Convert Sec $\theta$ and tan $\theta$ in  sin  $\theta$ and cos  $\theta$

We Know that ,

⇒  sec θ = $\frac{1}{Cos\theta}$

⇒  tan θ = $\frac{Sin\theta}{Cos\theta}$

⇒  $\frac{1}{Cos\theta}$ +  $\frac{Sin\theta}{Cos\theta}$ = p

Taking Cos θ Common

⇒ $\frac{1 + Sin\theta}{Cos\theta}= p$

Squaring both the Sides

⇒ $(\frac{1 + Sin\theta}{Cos\theta})^2= (p)^2$

⇒ $\frac{1 +\ Sin^2\theta \ + 2Sin^2\theta}{Cos^2\theta}$ = $p^2$

⇒ ${1 +\ Sin^2\theta + 2 Sin\theta = p^2 * Cos^2\theta$

⇒ ${1 +\ Sin^2\theta + 2 Sin\theta = p^2 * ( 1 - Sin^2\theta )$

⇒ ${1 +\ Sin^2\theta + 2 Sin\theta = p^2- p^2 \ Sin^2\theta$

Let x is equal to Sin θ

⇒$1 + x^2 + 2x = p^2 - p^2 x^2$

⇒ $x^2 + p^2x^2 + 2x + 1 - p^2 = 0$

⇒ $(1 - p^2)x^2+ 2x + (1 - p)^2 = 0$

Since we obtained a quadratic equation we need to solve by Quadratic formula or any other method which is easier for the above quadratic equation

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

⇒ $\displaystyle x=\frac{-2\pm\sqrt{4-4(1 + p^2) (1 -p^2)}}{2(1 +p^2)}$

⇒  $\displaystyle x=\frac{-2\pm\sqrt{4} \sqrt{1 - (1 + p^2) (1 -p^2)}}{2(1 +p^2)}$

⇒$\displaystyle x=\frac{-1\pm\sqrt{ p^4}}{(1 + p^2)}$

⇒ $x = \frac{-1 \ \pm \ p^2}{(1+ p^2)}$

Put x = Sin θ

⇒  $Sin =\frac{-1 \ \pm \ p^2}{(1+ p^2)}$

For Taking Positive Sign

$Sin\theta = \frac{-1 + p^2}{(1 + p^2)}$

⇒  Sin θ = $\frac{p^2 - 1}{p^2 + 1}$

As we know ,

Cosec  θ = $\frac{1}{Sin\theta}$

⇒ $\frac{p^2 + 1}{p^2 - 1}$

For Taking Negative Sign

$Sin\theta= \frac{-1 - p^2}{(1 + p^2)}$

⇒   Sin θ = -1

As we know ,

Cosec  θ = $\frac{1}{Sin\theta}$

⇒ $\frac{1}{-1}$

⇒ -1