Question

PLEASE GIVE ANSWER ​

Answer 1

Step-by-step explanation:

LHS=

1−sinA

cosA

+

1+sinA

cosA

=

(1−sin

2

A)

cosA(1+sinA+1−sinA)

=

cos

2

A

2cosA

=

cosA

2

=2secA=RHS

Hence proved.

Answer 2

Given:-

• $\sf \dfrac{cos\theta}{1 + sin\theta} + \dfrac{1+sin\theta}{cos\theta} = 2sec\theta$

We have to prove it.

Formulae to be used:-

• (a + b)² = a² + b² + 2ab
• sin²θ + cos²θ = 1
• 1/cosθ = secθ

Proof:-

$\blue{\bigstar} \underline{\sf{\blue{L.H.S}}}$

$\sf \dfrac{cos \theta}{1 + sin \theta} + \dfrac{1 + sin \theta}{cos \theta}$

$\longrightarrow \sf \dfrac{cos {}^{2} \theta \: + {(1 + sin \theta)}^{2} }{(1 + sin \theta)(cos \theta)}$

Using (a + b)² = a² + b² + 2ab,

$\longrightarrow \sf \: \dfrac{cos {}^{2} \theta \: + 1 + sin {}^{2} \theta + 2sin \theta }{(1 + sin \theta)(cos \theta)}$

$\longrightarrow \sf \: \dfrac{(cos {}^{2} \theta \: + \: sin {}^{2} \theta)\: + 1 + 2sin \theta }{(1 + sin \theta)(cos \theta)}$

Using sin²θ + cos²θ = 1,

$\longrightarrow \sf \: \dfrac{1\: + 1 + 2sin \theta }{(1 + sin \theta)(cos \theta)}$

$\longrightarrow \sf \: \dfrac{2 + 2sin \theta}{(1 + sin \theta)(cos \theta)}$

$\longrightarrow \sf \: \dfrac{2 \cancel{(1 + sin \theta)}}{ \cancel{(1 + sin \theta)}(cos \theta)}$

$\longrightarrow \sf \: \dfrac{2}{cos \theta}$

$\longrightarrow \sf \: 2 \times \dfrac{1}{cos \theta}$

Since 1/cosθ = secθ,

$\longrightarrow \underline{ \underline{ \sf{ \green{ \: \: 2sec \theta \: \: }}}}$

$\leadsto \pink{\bigstar} \sf{\pink{L.H.S}}$

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$