Question

Prove that (sin theta+1+cos theta) (sin theta-1+cos theta) sec theta cosec theta

Answer 1

Answer:

I think your question is not all right

Step-by-step explanation:

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Answer 2

Correct Question :-

• Prove that (sin theta+1+cos theta) (sin theta-1+cos theta) sec theta cosec theta = 2

Given :-

• $\mathsf{ (sin\theta + cos\theta + 1)(sin\theta + cos\theta - 1).({sec}\theta.co{sec}\theta)}$=2

To Prove :-

• Value of $\mathsf{ (sin\theta + cos\theta + 1)(sin\theta + cos\theta - 1).({sec}\theta.co{sec}\theta)}$= 2

Proof :-

$\qquad$L.H.S

$\mathsf\pink{{ (sin\theta + cos\theta + 1)(sin\theta + cos\theta - 1).({sec}\theta.co{sec}\theta)}}$

$\qquad$❏$\sf\red{\boxed{\sf{(a + b)(a - b) = a² - b²}}}$

$\mathsf{ \: \: \: \: \: \: \: \: \: \:\::\implies [(sin\theta + cos\theta)^2 - (1)^2].[{sec}\theta.co{sec}\theta]}$

$\mathsf{ \: \: \: \: \: \: \: \: \: \:\::\implies [sin^2\theta + cos^2\theta + 2.sin\theta.cos\theta - 1].[{sec}\theta.co{sec}\theta]}$

$\qquad$❏ $\sf\red{\boxed{\sf{sin²θ + cos²θ = 1}}}$

$\mathsf{ \: \: \: \: \: \: \: \: \: \:\::\implies [1 + 2.sin\theta.cos\theta - 1].[{sec}\theta.co{sec}\theta]}$

$\mathsf{ \: \: \: \: \: \: \: \: \: \:\::\implies [2.sin\theta.cos\theta].[{sec}\theta.co{sec}\theta]}$

$\mathsf{ \: \: \: \: \: \: \: \: \: \:\::\implies [2.sin\theta.co{sec}\theta].[{cos}\theta.{sec}\theta]}$

$\qquad$❏$\;\;\textsf{ \red{\boxed{\mathsf{{sec}\theta = \dfrac{1}{cos\theta}}}}}$

$\qquad$❏$\;\;\textsf{\red{ \boxed{\mathsf{co{sec}\theta = \dfrac{1}{sin\theta}}}}}$

$\mathsf{ \: \: \: \: \: \: \: \: \: \:\::\implies 2\bigg(\cancel{sin\theta} \times \dfrac{1}{\cancel{sin\theta}}\bigg).\bigg(\cancel{cos\theta} \times \dfrac{1}{\cancel{cos\theta}}\bigg)}$

$\mathsf{ \: \: \: \: \: \: \: \: \: \:\:\pink{:\implies 2}}$

$\qquad$R.H.S

$\therefore\:\underline{\textsf{ \textbf{Proved..!}}}.$