Question

एका पेटीत 64 आंबे याप्रमाणे 1200पेट्या होत्या.ते सर्व आंबे एकत्र करून नवीन 1280 पेट्या तयार केल्या; तर प्रत्येक पेटीत किती आंबे असतील? 
​

Answer 1

Answer:

Question

Prove that :

\sf\dfrac{\cot\theta+\csc\theta}{\sin\theta+\tan\theta}=\cot\theta\times\csc\theta

sinθ+tanθ

cotθ+cscθ

=cotθ×cscθ

Solution :

We have to prove that :

\rm\dfrac{\cot\theta+\csc\theta}{\sin\theta+\tan\theta}=\cot\theta\times\csc\theta

sinθ+tanθ

cotθ+cscθ

=cotθ×cscθ

Now ,If we convert LHS tanθ and cotθ in in terms of sinθ and cosθ, And Manipulate. We'll get answer.

Let's Try something Different :

First solve LHS

\bf\blue{LHS}LHS

\sf=\dfrac{\cot\theta+\csc\theta}{\sin\theta+\tan\theta}=

sinθ+tanθ

cotθ+cscθ

Multiply , Numerator and denominator by cotθ×cosec θ . Then ,

\sf=\dfrac{\cot\theta+\csc\theta}{\sin\theta+\tan\theta}\times\dfrac{\csc\theta\times\cot\theta}{\csc\theta\times\cot\theta}=

sinθ+tanθ

cotθ+cscθ

×

cscθ×cotθ

cscθ×cotθ

\sf=[\cot\theta\times\csc\theta]\times\dfrac{\cot\theta+\csc\theta}{\sin\theta+\tan\theta}\times\dfrac{1}{\csc\theta\times\cot\theta}=[cotθ×cscθ]×

sinθ+tanθ

cotθ+cscθ

×

cscθ×cotθ

1

\sf=[\cot\theta\times\csc\theta]\times\dfrac{\cot\theta+\csc\theta}{(\sin\theta+\tan\theta)(\csc\theta\times\cot\theta)}=[cotθ×cscθ]×

(sinθ+tanθ)(cscθ×cotθ)

cotθ+cscθ

\sf=[\cot\theta\times\csc\theta]\times\dfrac{\cot\theta+\csc\theta}{\sin\theta\csc\theta\cot\theta+\tan\theta\cot\csc\theta}=[cotθ×cscθ]×

sinθcscθcotθ+tanθcotcscθ

cotθ+cscθ

\sf=[\cot\theta\times\csc\theta]\times\dfrac{\cot\theta+\csc\theta}{\cot\theta+\csc\theta}=[cotθ×cscθ]×

cotθ+cscθ

cotθ+cscθ

\sf=\cot\theta\csc\theta=cotθcscθ

\bf\green{=RHS}=RHS

Hence, Proved !