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Translate into english Or Gujaratiनमः शम्भवाय च मयोभवाय च नमः शङ्कराय च मयस्कराय च नमः शिवाय च शिवतराय च |​

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Answer:

Answer:

▪ To Prove :-

\mathtt{ \dfrac{ cos \theta}{1 - tan \theta} + \dfrac{ sin \theta}{1 - cot \theta} = cos \theta + sin \theta}

1−tanθ

cosθ

+

1−cotθ

sinθ

=cosθ+sinθ

-------------------------------

▪ Formulae Used :-

\begin{gathered}\maltese \: \: \bf tan\theta=\dfrac{sin\theta}{cos\theta} \\ \\ \maltese \: \: \:\bf cot\theta=\dfrac{cos\theta}{sin\theta} \\ \\\maltese \: \: \:\bf a^2-b^2=(a+b)(a-b)\end{gathered}

✠tanθ=

cosθ

sinθ

✠cotθ=

sinθ

cosθ

✠a

2

−b

2

=(a+b)(a−b)

-------------------------------

▪ Proof :-

\begin{gathered}\large\mathcal{LHS} :\\ \\ \mathtt{ \dfrac{ cos \theta}{1 - tan \theta} + \dfrac{ sin \theta}{1 - cot \theta} } \\ \\ = \: \mathtt{ \dfrac{cos \theta}{1 - \dfrac{sin\theta}{cos\theta } } + \dfrac{sin\theta}{1 - \dfrac{cos\theta}{sin\theta} } } \\ \\ \: = \mathtt{ \frac{cos {}^{2} \theta}{cos\theta - sin\theta} + \frac{ {sin}^{2} \theta}{sin\theta - cos\theta} } \\ \\ \: = \mathtt{ \frac{ {cos}^{2}\theta - {sin}^{2} \theta }{cos\theta - sin\theta} } \\ \\ \: = \mathtt{ \frac{(cos\theta - sin\theta)(cos\theta + sin\theta)}{cos\theta - sin\theta} } \\ \\ \: = \mathtt{cos \theta + sin\theta }=\large\mathcal{RHS}\end{gathered}

LHS:

1−tanθ

cosθ

+

1−cotθ

sinθ

=

1−

cosθ

sinθ

cosθ

+

1−

sinθ

cosθ

sinθ

=

cosθ−sinθ

cos

2

θ

+

sinθ−cosθ

sin

2

θ

=

cosθ−sinθ

cos

2

θ−sin

2

θ

=

cosθ−sinθ

(cosθ−sinθ)(cosθ+sinθ)

=cosθ+sinθ=RHS

\huge\color{magenta} \red{\bigstar}\:\: \mathfrak{\text{H}ence\:\:Proved}\:\: \red{\bigstar}★HenceProved★

-------------------------------

\begin{gathered} \large\bf Fundamental \: Trigonometric \\ \Large \bf Identities \\ \\ \maltese \: \: \:\sin^2\theta + \cos^2\theta=1 \\ \\ \maltese \: \: \: 1+\tan^2\theta = \sec^2\theta \\ \\\maltese \: \: \: 1+\cot^2\theta = \text{cosec}^2 \, \theta\end{gathered}

FundamentalTrigonometric

Identities

✠sin

2

θ+cos

2

θ=1

✠1+tan

2

θ=sec

2

θ

✠1+cot

2

θ=cosec

2

θ

-------------------------------

ok

hope it helps you...
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