Question

If 15 tan^2 Q+ 4 sec ^2 Q= 23, then find the value of ( sec Q + cosec Q )^2- sin^2 Q

Answer 1

Solution :-

→ 15tan²Q + 4sec²Q = 23

using tan²A = (sec²A - 1) ,

→ 15( sec²Q -1) + 4sec²Q = 23

→ 15sec²Q + 4sec²Q - 15 = 23

→ 19sec²Q = 23 + 15

→ 19sec²Q = 38

→ sec²Q = 2

→ secQ = (√2/1) = Hypotenuse/Base.

So ,

→ Perpendicular = √(Hypotenuse² - Base²)

→ Perpendicular = √{(√2)² - 1²}

→ Perpendicular = √(2 - 1)

→ Perpendicular = √1 = 1 .

Therefore,

→ cosecQ = Hypotenuse / Perpendicular = √2/1 = √2.

→ sinQ = Perpendicular / Hypotenuse = 1/√2 .

Hence,

→ (secQ + cosecQ)² - sin²Q

→ (√2 + √2)² - (1/√2)²

→ (2√2)² - (1/2)

→ 8 - (1/2)

→ (15/2) (Ans.)

Answer 2

Question:-

If 15tan²Q + 4sec²Q = 23, find the value of (secQ + cosecQ)² - sin²Q

Formulas Used:-

• tan²Q + 1 = sec²Q
• (Base)² + (Perpendicular)² = (Hypotenuse)²

Answer:-

15tan²Q + 4sec²Q = 23

=> 15tan²Q + 4(tan²Q + 1) = 23

=> 15tan²Q + 4tan²Q + 4 = 23

=> 19tan²Q = 19

=> tan²Q = 1

=> tanQ = 1 = 1/1 = (Perpendicular) / (Base)

We are having the value of Perpendicular as 1 and Base as 1

So, (Base)² + (Perpendicular)² = (Hypotenuse)²

=> 1² + 1² = (Hypotenuse)²

=> (Hypotenuse)² = 2

=> Hypotenuse = √2

So,

secQ = (Hypotenuse) / (Base) = √2 / 1 = √2

cosecQ = (Hypotenuse) / (Perpendicular) = √2 / 1 = √2

sinQ = (Perpendicular) / (Hypotenuse) = 1/√2

So,

(secQ + cosecQ)² - sin²Q

=> ( √2 + √2 )² - (1/√2)²

=> (2√2)²- (1/2)

=> 8 - 1/2

=> 15/2

Ans. 15/2