Question

Solve it urgently [Trigonometry Class X]

Answer 1

(i) $\frac{cos 58}{sin (90-58)} + \frac{sin 22}{cos (90-22)} - \frac{cos 38.cosec(90-38)}{tan 18 . tan 35.tan (90-18).tan (90-35).tan 60}$
= $\frac{cos 58}{cos 58} + \frac{sin 22}{sin 22} - \frac{cos 38.sec38}{tan 18 . tan 35.cot18.cot 35. \sqrt{3} }$
= 1 + 1 - $\frac{1}{ \sqrt{3} }$
= 2 - 1/√3
= (2√3-1)/√3
= (2√3-1)/√3 × √3/√3
= (6-√3)/3

(ii) $= \frac{sin 15.cos (90-15) + cos 15.sin(90-15)}{tan 5. tan 30. tan (90-55).tan 55.tan (90-5) }$
$= \frac{sin 15.sin15+ cos 15.cos 15}{tan 5. tan 30.cot 55.tan 55.cot 5 }$
$= \frac{ sin^{2} 15+ cos^{2} 15}{tan 30}$
$= \frac{1}{\frac{1}{ \sqrt{3}}}$
=√3

(iii) $= \frac{3 cos 55}{7 sin (90-55)} - \frac{4 cos70.cosec(90-70)}{7(tan 5.tan 25.tan 45.tan (90-25).tan(90-5) }$
$= \frac{3 cos 55}{7 cos55} - \frac{4 cos70.sec 70}{7(tan 5.tan 25.tan 45.cot 25.cot 5}$
$= \frac{3}{7} - \frac{4}{7(tan 45)}$
$= \frac{3}{7} - \frac{4}{7(1)}$
$= \frac{3}{7} - \frac{4}{7}$
$= \frac{3-4}{7}$
= -1/7

{tan 30° = 1/√3
tan 45° = 1
sec ∅ = 1/cos ∅
sin ∅ = cos (90-∅)
Cos ∅ = sin (90-∅)
sin²@+cos²@ = 1}

Hope it helps

Answer 2

okk here is the solutions