tan \(\theta \) cosec²\(\theta \) - tan\(\theta \) is equal to 

(1 + tan \(\theta \) + sec\(\theta \)) (1 + cot\(\theta \) - cosec\(\theta \)) is equal to 

If the ratio of the height of a tower and the length of its shadow is \(\sqrt{3}: 1\), then the angle of elevation of the sun has measure

The angle of elevation of a cloud from a point h metres above a lake is \(\beta \). The angle of depression of its reflection in the lake is 45°. The height of location of the cloud from the lake is

The value of the expression [cosec (75^o + θ) - sec (15^o - θ) - tan (55^o + θ) + cot(35^o - θ] is ___________

prove that \(\frac { sinA }{ 1+cosA } =\frac { 1-cosA }{ sinA } \) 

prove that \(\frac { sec\theta }{ sin\theta } -\frac { sin\theta }{ cos\theta } =cot\theta \)

prove that \(\frac { sinA }{ secA+tanA-1 } +\frac { cosA }{ cosecA+cotA-1 } =1\)

The horizontal distance between two buildings is 140 m. The angle of depression of the top of the first building when seen from the top of the second building is 30° . If the height of the first building is 60 m, find the height of the second building.(\(\sqrt { 3 } \) = 1.732)

A man is watching a boat speeding away from the top of a tower. The boat makes an angle of depression of 60° with the man’s eye when at a distance of 200 m from the tower. After 10 seconds, the angle of depression becomes 45°. What is the approximate speed of the boat (in km / hr), assuming that it is sailing in still water ?(\(\sqrt { 3 } \) = 1.732)

calculate \(\angle \)BAC in the given triangles (tan 38.7° = 0.8011 )

From a point on the ground, the angles of elevation of the bottom and top of a tower fixed at the top of a 30m high building are \(45°\)and \(60°\) respectively. find the height of the tower. (\(\sqrt { 3 } =1.732\) )

calculate \(\angle \)BAC in the given triangles ( tan 69.4° = 2.6604 )

prove the following identity.
cot \(\theta \) + tan \(\theta \) = sec \(\theta \) cosec\(\theta \)

Prove that cot²A\(\left( \frac { secA-1 }{ 1+sinA } \right) \) + sec²A\(\left( \frac { sinA-1 }{ 1+secA } \right) \) = 0