Question

prove the identity sinA (1 + tanA ) + cosA (1 + cotA) = secA + cosecA​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{sin(A)\,(1+tan(A))\,+cos(A)\,(1+cot(A))}$

$\sf{=sin(A)\,\bigg\{1+\dfrac{sin(A)}{cos(A)}\bigg\}\,+cos(A)\,\bigg\{1+\dfrac{cos(A)}{sin(A)}\bigg\}}$

$\sf{=sin(A)\,\bigg\{\dfrac{cos(A)+sin(A)}{cos(A)}\bigg\}\,+cos(A)\,\bigg\{\dfrac{sin(A)+cos(A)}{sin(A)}\bigg\}}$

$\sf{=\dfrac{sin(A)}{cos(A)}\,\{sin(A)+cos(A)\}\,+\dfrac{cos(A)}{sin(A)}\,\{sin(A)+cos(A)\}}$

$\sf{=tan(A)\,\{sin(A)+cos(A)\}\,+cot(A)\,\{sin(A)+cos(A)\}}$

$\sf{=\{sin(A)+cos(A)\}\{tan(A)+cot(A)\}}$

$\sf{=\{sin(A)+cos(A)\}\,\bigg\{\dfrac{sin(A)}{cos(A)}+\dfrac{cos(A)}{sin(A)}\bigg\}}$

$\sf{=\{sin(A)+cos(A)\}\,\bigg\{\dfrac{sin^2(A)+cos^2(A)}{cos(A)\,sin(A)}\bigg\}}$

$\sf{=\{sin(A)+cos(A)\}\,\bigg\{\dfrac{1}{cos(A)\,sin(A)}\bigg\}}$

$\sf{=\dfrac{sin(A)+cos(A)}{cos(A)\,sin(A)}}$

$\sf{=\dfrac{sin(A)}{cos(A)\,sin(A)}+\dfrac{cos(A)}{cos(A)\,sin(A)}}$

$\sf{=\dfrac{1}{cos(A)}+\dfrac{1}{sin(A)}}$

$\sf{=sec(A)+cosec(A)}$