Question

(sin theta+1+cos theta)(sin theta -1+cos theta)sec theta.cosec theta=2​

Answer 1

Answer:

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Answer 2

$(\sin \theta+1+\cos \theta)(\sin \theta -1+\cos \theta)\sec \theta\cdot\csc \theta=2$ Proved.

Step-by-step explanation:

Consider the provided information.

$(\sin \theta+1+\cos \theta)(\sin \theta -1+\cos \theta)\sec \theta\cdot\csc \theta=2$

Consider the LHS.

$(\sin \theta+1+\cos \theta)(\sin \theta -1+\cos \theta)\sec \theta\cdot\csc \theta$

Rewrite the equation.

$((\sin \theta+\cos \theta)+1)((\sin \theta+\cos \theta)-1)\sec \theta\cdot\csc \theta$

Now use the formula: $a^2-b^2=(a+b)(a-b)$

$((\sin \theta+\cos \theta)^2-1^2)\sec \theta\cdot\csc \theta$

$(\sin^2 \theta+\cos^2 \theta+2\sin\theta\cos\theta-1)\sec \theta\cdot\csc \theta$

$(1+2\sin\theta\cos\theta-1)\sec \theta\cdot\csc \theta$ (∴ sin²θ+cos²θ=1)

$(2\sin\theta\cos\theta)\frac{1}{\cos\theta}\cdot\frac{1}{\sin\theta}=2$

Hence, proved LHS=RHS

#Learn more

Prove that

Sin80 -cos110 = root3sin50

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