Question

prove that :
.

class 11
​

Answer 1

Step-by-step explanation:

here your answer check it

Answer 2

Answer:

Given Question :

Prove that :

$\mathrm{\dfrac{tan\bigg(\dfrac{\pi}{2} -x\bigg)sec(\pi-x)sin(-x)}{sin(\pi+x)cot(2\pi-x)cosec\bigg(\dfrac{\pi}{2}-x\bigg)} = 1}$

Solution :

Given LHS is

$\longmapsto \mathrm{LHS = \dfrac{tan\bigg(\dfrac{\pi}{2} -x\bigg)sec(\pi-x)sin(-x)}{sin(\pi+x)cot(2\pi-x)cosec\bigg(\dfrac{\pi}{2}-x\bigg)} }$

We know that from the CAST diagram,

$\setlength{\unitlength}{7mm}\begin{picture}(0,0)\thicklines\put(0,0){\vector(1,0){6}}\put(0,0){\vector(-1,0){6}}\put(0,0){\vector(0,1){6}}\put(0,0){\vector(0,-1){6}}\put( -3,4){\sf\huge S}\put(3,4){\sf\huge A}\put( -3, - 2){\sf\huge T}\put(3, - 2){\sf\huge C}\put( -4.5,2.5){\sf\large Sin is positive}\put(1.5,2.5){\sf\large All are positive}\put( -4.5, - 3){\sf\large Tan is positive}\put(2, - 3){\sf\large Cos is positive}\put( -4.1,1.5){\sf Cos and Tan }\put( -4.1,1){\sf are negative}\put( -4.1, -4){\sf Sin and Cos}\put( -4.1, - 4.5){\sf are negative}\put(2.4, - 4){\sf Sin and Tan}\put(2.4, - 4.5){\sf are negative}\end{picture}$

$\boxed{\begin{array}{cc} \mathrm{tan\bigg(\dfrac{\pi}{2}-x\bigg) = cotx}\\\\\mathrm{sec(\pi-x) = -secx}\\\mathrm{sin(-x) = -sinx}\\\mathrm{sin(\pi+x) = -sinx}\\\mathrm{cot(2\pi-x) = -cotx}\\\mathrm{cosec\bigg(\dfrac{\pi}{2}-x\bigg) = secx}\end{array}}$

So, applying this in the above LHS, we get

$\implies \mathrm{\dfrac{cotx\: (-secx)(-sinx)}{(-sinx)(-cotx)\, secx} }$

Crossing out the like terms, we get,

$\implies \mathrm{\dfrac{\cancel{cotx}\: \bcancel{(-secx)}(-sinx)}{(-sinx)(-\cancel{cotx})\, \bcancel{secx}}}$

=> 1

=> RHS

Thus, LHS = RHS

Hence Proved!!

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Hope it helps!!