Question

Prove that
sinA + cosA/sinA - cosA +sinA-cosA/sinA+cosA = 2/ sin^2A-cos^2A = 2/2sin^2-A- 1 = 2/1-2cos^2A


question 27
please help
tomorrow is my exam

Answer 1

Answer:

$\frac{sinA+cosA}{sinA-cosA}+\frac{sinA-cosA}{sinA+cosA}\\=\frac{2}{sin^{2}A-cos^{2}A}\\=\frac{2}{2sin^{2}A-1}\\=\frac{2}{1-2cos^{2}A}$

Step-by-step explanation:

$\frac{sinA+cosA}{sinA-cosA}+\frac{sinA-cosA}{sinA+cosA}\\=\frac{(sinA+cosA)^{2}+(sinA-cosA)^{2}}{(sinA-cosA)(sinA+cosA)}$

$=\frac{2(sin^{2}A+cos^{2}A)}{sin^{2}A-cos^{2}A}$

/* By algebraic identities:

1) (a+b)²+(a-b)² = 2(a²+b²)

2) (a-b)(a+b) = a²-b²

and

By Trigonometric identity:

sin²A+cos²A = 1 */

*/

$=\frac{2}{sin^{2}A-cos^{2}A}---(1)$

$=\frac{2}{sin^{2}A-(1-sin^{2}A)}$

/* By Trigonometric identity:

cos²A = 1-sin²A*/

$=\frac{2}{sin^{2}A-1+sin^{2}A}\\=\frac{2}{2sin^{2}A-1}---(2)$

$=\frac{2}{2(1-cos^{2}A)-1}\\=\frac{2}{2-2cos^{2}A-1}\\=\frac{2}{1-2cos^{2}A}--(3)--(3)$

Therefore,

$\frac{sinA+cosA}{sinA-cosA}+\frac{sinA-cosA}{sinA+cosA}\\=\frac{2}{sin^{2}A-cos^{2}A}\\=\frac{2}{2sin^{2}A-1}\\=\frac{2}{1-2cos^{2}A}$

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Answer 2

Answer:  The proof is given below.

Step-by-step explanation:  We are given to prove the following trigonometric equalities :

$\dfrac{\sin A+\cos A}{\sin A-\cos A}+\dfrac{\sin A-\cos A}{\sin A+\cos A}=\dfrac{2}{\sin^2A-\cos^2A}=\dfrac{2}{2\sin^2A-1}=\dfrac{2}{1-\cos^2A}.$

We will be using the following trigonometric identities :

$\sin^2\theta+\cos^2\theta=1,\\\\\sin^2\theta=1-\cos^2\theta,\\\\\cos^2\theta=1-\sin^2\theta.$

We have

$\dfrac{\sin A+\cos A}{\sin A-\cos A}+\dfrac{\sin A-\cos A}{\sin A+\cos A}\\\\\\=\dfrac{(\sin A+\cos A)^2+(\sin A-\cos A)^2}{(\sin A-\cos A)(\sin A+\cos A)}\\\\\\=\dfrac{(\sin A)^2-\sin A\cos A+(\cos A)^2+(\sin A)^2+\sin A\cos A+(\cos A)^2}{(\sin A)^2-(\cos A)^2}\\\\\\=\dfrac{2\sin^2A+2\cos^2A}{\sin^2A-\cos^2A}\\\\\\=\dfrac{2(\sin^2A+\cos^2A)}{\sin^2A-\cos^2A}\\\\\\=\dfrac{2\times1}{\sin^2A-\cos^2A}\\\\\\=\dfrac{2}{\sin^2A-\cos^2A}\\\\\\=\dfrac{2}{\sin^2A-(1-\sin^2A)}\\\\\\=\dfrac{2}{2\sin^2A-1}\\\\\\=\dfrac{2}{2(1-\cos^2A)-1}=\dfrac{2}{1-2\cos^2A}.$

Hence proved.