The value of the expression cosec(75°+θ)-sec(15°-θ)-tan(55°+θ)+cot(35°-θ) is
-1
0
1
32
In this question we use the formula cosec(90-θ)=secθ,cot(90-θ)=tanθ
Here we use cosec(75°+θ)=cosec(90-(15-θ)),cot(35°-θ)=cot(90-(55+θ))
cosec(75°+θ)-sec(15°-θ)-tan(55°+θ)+cot(35°-θ)=cosec(90-(15-θ))-sec(15°-θ)-tan(55°+θ)+cot(90-(55°+θ))=sec(15°-θ)-sec(15°-θ)-tan(55°+θ)+tan(55°+θ)=0
Hence option(B) is the correct answer and all other options are incorrect answers.
Prove that:
(i) sin(70∘+θ)−cos(20∘−θ)=0
(ii) tan(55∘−θ)−cot(35∘+θ)=0
(iii) cosec(67∘+θ)−sec(23∘−θ)=0
(iv) cosec(65∘+θ)−sec(25∘−θ)−tan(55∘−θ)+cot(35∘+θ)=0
(v) sin(50∘+θ)−cos(40∘−θ)+tan 1∘ tan 10∘ tan 80∘ tan 89∘=1