if cos a=15 then the value of (1+sina)(1-sina)\(1+cosa)(1-cosa)
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if cos a=15 then the value of (1+sina)(1-sina)\(1+cosa)(1-cosa)
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To find the value of \(\frac{{(1+\sin a)(1-\sin a)}}{{(1+\cos a)(1-\cos a)}}\), you can use trigonometric identities.
To find the value of \(\frac{{(1+\sin a)(1-\sin a)}}{{(1+\cos a)(1-\cos a)}}\), you can use trigonometric identities. Given that \(\cos a = 15\), we can calculate \(\sin a\) using the Pythagorean identity for sine:
To find the value of \(\frac{{(1+\sin a)(1-\sin a)}}{{(1+\cos a)(1-\cos a)}}\), you can use trigonometric identities. Given that \(\cos a = 15\), we can calculate \(\sin a\) using the Pythagorean identity for sine:\[\sin^2 a + \cos^2 a = 1\]
To find the value of \(\frac{{(1+\sin a)(1-\sin a)}}{{(1+\cos a)(1-\cos a)}}\), you can use trigonometric identities. Given that \(\cos a = 15\), we can calculate \(\sin a\) using the Pythagorean identity for sine:\[\sin^2 a + \cos^2 a = 1\]\[\sin^2 a + 15^2 = 1\]
To find the value of \(\frac{{(1+\sin a)(1-\sin a)}}{{(1+\cos a)(1-\cos a)}}\), you can use trigonometric identities. Given that \(\cos a = 15\), we can calculate \(\sin a\) using the Pythagorean identity for sine:\[\sin^2 a + \cos^2 a = 1\]\[\sin^2 a + 15^2 = 1\]\[\sin^2 a = 1 - 225\]
To find the value of \(\frac{{(1+\sin a)(1-\sin a)}}{{(1+\cos a)(1-\cos a)}}\), you can use trigonometric identities. Given that \(\cos a = 15\), we can calculate \(\sin a\) using the Pythagorean identity for sine:\[\sin^2 a + \cos^2 a = 1\]\[\sin^2 a + 15^2 = 1\]\[\sin^2 a = 1 - 225\]\[\sin^2 a = -224\]
To find the value of \(\frac{{(1+\sin a)(1-\sin a)}}{{(1+\cos a)(1-\cos a)}}\), you can use trigonometric identities. Given that \(\cos a = 15\), we can calculate \(\sin a\) using the Pythagorean identity for sine:\[\sin^2 a + \cos^2 a = 1\]\[\sin^2 a + 15^2 = 1\]\[\sin^2 a = 1 - 225\]\[\sin^2 a = -224\]Since \(\sin^2 a\) cannot be negative, it means that there is no real value of \(\sin a\) for which \(\cos a = 15).
To find the value of \(\frac{{(1+\sin a)(1-\sin a)}}{{(1+\cos a)(1-\cos a)}}\), you can use trigonometric identities. Given that \(\cos a = 15\), we can calculate \(\sin a\) using the Pythagorean identity for sine:\[\sin^2 a + \cos^2 a = 1\]\[\sin^2 a + 15^2 = 1\]\[\sin^2 a = 1 - 225\]\[\sin^2 a = -224\]Since \(\sin^2 a\) cannot be negative, it means that there is no real value of \(\sin a\) for which \(\cos a = 15).Therefore, the expression \(\frac{{(1+\sin a)(1-\sin a)}}{{(1+\cos a)(1-\cos a)}}\) is undefined in this case.
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