If sin\( \theta \) = K find tan\(\theta\), 0° \(\leq\) \(\theta\) \(\leq\) 90°.
1-K
\( \frac{k}{k - 1} \)
\( \frac{k}{\sqrt{1 - k^2}} \)
\( \frac{k}{1 - k} \)
\( \frac{k}{\sqrt{ k^2 - 1}} \)
Correct answer is C
\(\sin \theta = \frac{k}{1}\)
\(\implies 1^2 = k^2 + adj^2\)
\(adj = \sqrt{1 - k^2}\)
\(\therefore \tan \theta = \frac{k}{\sqrt{1 - k^2}}\)
If the 2nd and 5th terms of a G.P are 6 and 48 respectively, find the sum of the first for term
-45
-15
15
33
45
Correct answer is E
T\(_{2}\) = ar = 6
T\(_{5}\) = ar\(^{4}\) = 48
\(\frac{T_5}{T_2}\) = \(\frac{ar^{4}}{ar}\) = \(\frac{48}{6}\)
= r\(^{3}\) = 8
⇒ r = 2
S\(_{n}\) = \(\frac{a((r^n) - 1)}{r - 1}\)
S\(_{4}\) = \(\frac{a((r^4) - 1)}{r - 1}\)
S\(_{4}\) = \(\frac{3((2^4) - 1)}{2 - 1}\)
= 3(16 -1)
= 45