Sinθ × cosθ × secθ × cotθ ⇒ sinθ x cosθ x (1/cosθ) x (cosθ/sinθ) ⇒ sinθ x 1 x (cosθ/sinθ) ⇒ cosθ ...

You visited us 5 times! Enjoying our articles? Unlock Full Access!

Trigonometric Ratios of Standard Angles

What is sinθ×...

Question

What is sin $θ$ $\times$ cos $θ$ $\times$ sec $θ$ $\times$ cot $θ$ ?

Open in App

Solution

sin $θ$ $\times$ cos $θ$ $\times$ sec $θ$ $\times$ cot $θ$
$\Rightarrow$ sin $θ$ x cos $θ$ x (1/cos $θ$ ) x (cos $θ$ /sin $θ$ )
$\Rightarrow$ sin $θ$ x 1 x (cos $θ$ /sin $θ$ )
$\Rightarrow$ cos $θ$

Suggest Corrections

Similar questions

\frac{sin θ + cos θ}{sin θ - cos θ} + \frac{sin θ - cos θ}{sin θ + cos θ} = \frac{2 {sec}^{2} θ}{{tan}^{2} θ - 1}

\frac{sin θ}{cos θ} + \frac{cos θ}{sin θ} = 2

0 < θ < 90^{\circ}

θ

\frac{sin θ + cos θ}{sin θ - cos θ} + \frac{sin θ - cos θ}{sin θ + cos θ} = \frac{2}{2 {sin}^{2} θ - 1} = \frac{2}{1 - 2 {cos}^{2} θ}

Prove the following trigonometric identities:

Prove that:

(i) $\sqrt{\frac{s e c θ - 1}{s e c θ + 1}} + \sqrt{\frac{s e c θ + 1}{s e c θ - 1}} = 2 c o s e c θ$

(ii) $\sqrt{\frac{1 + s i n θ}{1 - s i n θ}} + \sqrt{\frac{1 - s i n θ}{1 + s i n θ}} = 2 c o s e c θ$

(iii) $\sqrt{\frac{1 + c o s θ}{1 - c o s θ}} + \sqrt{\frac{1 - c o s θ}{1 + c o s θ}} = 2 c o s e c θ$

(iv) $\frac{s e c θ - 1}{s e c θ + 1} = {(\frac{s i n θ}{1 + c o s θ})}^{2}$

\frac{sin θ - cos θ + 1}{sin θ + cos θ + 1} = \frac{1}{sec θ - tan θ}

{sec}^{2} θ = 1 + {tan}^{2} θ

Join BYJU'S Learning Program