Trigonometry Questions and Answers

We know that

$sec^2theta - tan^2 theta = 1$
and $sec theta + tan theta = 2$(given)    ...(i)

then, $frac{sec^2 theta - tan^2 theta}{sec theta + tan theta} = frac{1}{2}$

$therefore sec theta - tan theta = frac{1}{2}$ ...(ii)

Subtracting (ii) from (i)

$2 tan theta = frac{3}{2} Rightarrow tan theta = frac{3}{4}$

$cos^2 25degree = cos^2 (90degree - 25degree)$

$= sin^2 65degree$

$therefore cos^2 25degree + cos^2 65degree = sin^2 65degree + cos^2 65degree$

$= 1 (because sin^2 theta + cos^2 theta = 1)$

Given, $dfrac{cos^2theta - 3costheta + 2}{sin^2theta} = 1$

$Rightarrow cos^2theta - 3costheta + 2 = sin^2theta = 1 - cos^2 theta$

$Rightarrow 2 cos^2 theta - 3 cos theta + 1 = 0$

$Rightarrow 2 cos^2 theta - 2 cos theta - cos theta + 1 = 0$

$Rightarrow (2 cos theta - 1)(cos theta - 1) = 0$ either $cos theta = 1 quad text{ or } (2 cos theta - 1) = 0$ but $cos theta ne 1$

$therefore cos theta = 1$

$therefore cos theta = frac{1}{2} = cos 60degree$

$therefore theta = 60^degree$

$sin theta = frac{text{a} + 2text{b}}{2text{b}} = frac{text{a}}{2text{b}} + 1 gt 1$

$[because text{ a } text{ and } text{ b } text{ are the } therefore frac{text{a}}{2text{b}} text{ in the} ]$

But $sin theta$ lies between -1 and 1

$therefore$ Not possible

$frac{h}{100} = tan 45degree$

$Rightarrow h = 100 text{ m }$

$frac{cos^3 theta - sin^3 theta}{cos theta - sin theta}$

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

$= 1 + frac{2 sin theta cos theta}{2} (because sin^2 theta + cos^2 theta = 1)$

$= 1 + frac{sin 2 theta}{2}$

tan A + cot A = 2

$therefore frac{sin A}{cos A} + frac{cos A}{sin A} =2 Rightarrow sin^2 A + cos^2 A$

$=2 sin A cos A Rightarrow 1 = sin 2A Rightarrow sin 2A$

$=sin 90degree Rightarrow 2A = 90 Rightarrow A = cos 45degree$

$= sec A = sec 45 = sqrt2$