Practice30 Create account

If $A + B = 90$ °, then (tanA·tanB + tanA·cotB)/(sinA·secB) − si $n^{2}$ $B$ /co $s^{2}$ A simplifies to:

Competition practiceTrigonometry

Ase $c^{2}$ A
B1
Cco $t^{2}$ A
Dta $n^{2}$ A

Step-by-Step Solution

With

B = 90

° − A the numerator is 1 + ta

n^{2}

A

= se

c^{2}

A and the denominator sinA·cosecA = 1; also si

n^{2}

B

/co

s^{2}

A = 1

, leaving se

c^{2}

A - 1

= ta

n^{2}

A.

Answer: D) ta

n^{2}

A

Master this topic in 30 minutes a day.

Don't just memorise the answer — build a daily practice habit.

Practice 5 Similar Questions Free →

All Trigonometry practice questions

More Trigonometry practice

The angle of elevation of the top of a hill from the foot of a 20 m tower is 60°, and…If 7sin²θ + 3cos²θ = 4 and θ is acute, then tanθ equals:If a·cosθ − b·sinθ = c, then a·sinθ + b·cosθ equals:If tanθ = 3/4 and θ is acute, the value of (sinθ + cosθ)/(sinθ − cosθ) is:If sin⁴θ + cos⁴θ = 5/8 and θ is acute, then sinθ·cosθ equals: