Question

show that root of sec ² theeta +Cosec ²theeta=tantheeta +cottheeta​

Answer 1

Question :-

Prove that,

$\star \sqrt{ { \sec }^{2} \theta + { \csc}^{2} \theta } = \tan \theta + \cot \theta$

Solution :-

Taking left hand side ,

$\implies \: \sqrt{ { \sec }^{2} \theta + { \csc}^{2} \theta } \: \\ \\ \because \: { \sec }^{2} \theta - { \tan }^{2} \theta = 1 \\ \\ \: \: \: \: \: { \csc }^{2} \theta + { \cot }^{2} \theta \: = 1 \\ \therefore \: \\ \implies \: \sqrt{1 + { \tan }^{2} \theta + 1 + { \cot }^{2} \theta } \\ \\ \implies \: \sqrt{ { \tan}^{2} \theta \: + { \cot }^{2} \theta + 2 } \\ \\ \because \: { \tan }^{2} \theta \: { \cot }^{2} \theta = 1 \\ \\ \therefore \\ \\ \small \implies \: \sqrt{ { \tan}^{2} \theta \: + { \cot }^{2} \theta + 2 { \tan }^{2} \theta \: { \cot }^{2} \theta } \: \\ \\ \implies \: \sqrt{ {( \tan \theta + \cot \theta)}^{2} } \\ \\ \implies \: \tan \theta + \cot \theta\:$

LHS = RHS

Hence proved .

Answer 2

Step-by-step explanation:

Question :-

Prove that,

\begin{lgathered}\star \sqrt{ { \sec }^{2} \theta + { \csc}^{2} \theta } = \tan \theta + \cot \theta \\\end{lgathered}

⋆

sec

2

θ+csc

2

θ

=tanθ+cotθ

Solution :-

Taking left hand side ,

\begin{lgathered}\implies \: \sqrt{ { \sec }^{2} \theta + { \csc}^{2} \theta } \: \\ \\ \because \: { \sec }^{2} \theta - { \tan }^{2} \theta = 1 \\ \\ \: \: \: \: \: { \csc }^{2} \theta + { \cot }^{2} \theta \: = 1 \\ \therefore \: \\ \implies \: \sqrt{1 + { \tan }^{2} \theta + 1 + { \cot }^{2} \theta } \\ \\ \implies \: \sqrt{ { \tan}^{2} \theta \: + { \cot }^{2} \theta + 2 } \\ \\ \because \: { \tan }^{2} \theta \: { \cot }^{2} \theta = 1 \\ \\ \therefore \\ \\ \small \implies \: \sqrt{ { \tan}^{2} \theta \: + { \cot }^{2} \theta +

⟹

sec

2

θ+csc

2

θ

∵sec

2

θ−tan

2

θ=1

csc

2

θ+cot

2

θ=1

∴

⟹

1+tan

2

θ+1+cot

2

θ

⟹

tan

2

θ+cot

2

θ+2

∵tan

2

θcot

2

θ=1

∴

⟹

tan

2

θ+cot

2

θ+2tan

2

θcot

2

θ

⟹

(tanθ+cotθ)

2

⟹tanθ+cotθ

LHS = RHS

Hence proved

Thankuu mate ❤✌