Prove the following trigonometric identities- 1-sin-cos-cos-1-sin-2-sec-

We know that cos^2θ+ sin^2θ = 1 sec θ = 1/cos θ LHS = 1+sinθ/cosθ+cosθ/1+sinθ = (1+sinθ) × (1+sinθ) + cosθ× cosθ/cosθ(1+sinθ) = (1+sinθ)^2 + cos^2θ/cosθ(1+sinθ) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Trigonometric Identities

Prove the fol...

Question

Prove the following trigonometric identities:

$\frac{1 + s i n θ}{c o s θ} + \frac{c o s θ}{1 + s i n θ} = 2 s e c θ$

Open in App

Solution

We know that $c o s^{2} θ + s i n^{2} θ = 1$

$s e c θ = \frac{1}{c o s θ}$

$L H S = \frac{1 + s i n θ}{c o s θ} + \frac{c o s θ}{1 + s i n θ}$

$= \frac{(1 + s i n θ) \times (1 + s i n θ) + c o s θ \times c o s θ}{c o s θ (1 + s i n θ)}$

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

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

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

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

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

$= \frac{2}{c o s θ}$

$= 2 s e c θ = R H S$

Suggest Corrections

Similar questions

Prove the following trigonometric identities.
(i) $\frac{1 + cos θ + sin θ}{1 + cos θ - sin θ} = \frac{1 + sin θ}{cos θ}$

(ii) $\frac{sin θ - cos θ + 1}{sin θ + cos θ - 1} = \frac{1}{sec θ - tan θ}$

(iii) $\frac{cos θ - sin θ + 1}{cos θ + sin θ - 1} = cosec θ + cot θ$

(iv) $(sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ$

Q. Prove that

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

, using the identity

{sec}^{2} θ = 1 + {tan}^{2} θ

Q. IS LHS=RHS ?

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

Say yes or no.

Join BYJU'S Learning Program

Prove the following trigonometric identities: 1+sin θcos θ+cos θ1+sin θ=2 sec θ

We know that cos2θ+sin2θ=1 sec θ=1cos θ LHS=1+sinθcosθ+cosθ1+sinθ =(1+sinθ)×(1+sinθ)+cosθ×cosθcosθ(1+sinθ) =(1+sinθ)2+cos2θcosθ(1+sinθ) =1+2sinθ+sin2θ+cos2θcosθ(1+sinθ) =1+2sinθ+1cosθ(1+sinθ) =2+2sinθcosθ(1+sinθ) =2(1+sinθ)cosθ(1+sinθ) =2cosθ =2secθ=RHS

Prove the following trigonometric identities:

$\frac{1 + s i n θ}{c o s θ} + \frac{c o s θ}{1 + s i n θ} = 2 s e c θ$