Question

⇒ Kindly explain your answer
⇒ Avoid spamming​

Answer 1

Answer:

So, the option is b and c.

Explaination:

⇒ sec²A  + sec²B = sec²A . sec²B

⇒ sec²A = sec²A . sec²B - sec²B

⇒ sec²A = sec²B (sec²A - 1)

    (∵ sec²x - 1 = tan²x)

⇒ sec²A = sec²B . tan²A

⇒ sec²A/tan²A = sec²B

⇒ (1/cos²A) / (sin²A/cos²A)

⇒ 1/sin²A = sec²B

⇒ sin²A = cos²B

⇒ sin²A = sin²(90 - B)

⇒ sin²A - sin²(90 - B) = 0

⇒ (sinA + sin(90 - B))(sinA - sin(90-B)) = 0

∵ A and B are below 180° and in both quadrant sin is positive, so,

⇒ sinA - sin(90 - B) = 0

⇒ sinA = sin(90 - B)

so, ⇒ A = nπ + (-1)^n(π/2 - B)

(i)  put n = 0

⇒ A = π/2 - B

⇒ A + B = π/2

(ii) put n = 1

⇒ A = π - (π/2 - B)

⇒ A = π/2 + B

(iii) put n = 2

⇒ A = 2π + (π/2 - B)

⇒ A + B = 5π/2

This is not exist in triangle...

so, the angle has right angle and obtuse angle.

Answer 2

$\large\underline{\bold{Given \:Question - }}$

In triangle ABC, if

$\sf \: {sec}^{2}A + {sec}^{2}B = {sec}^{2}A {sec}^{2}B \: then \: \triangle \: ABC \: is \:$

$\large\underline{\sf{Solution-}}$

Identities Used :-

$\boxed{ \red{ \sf \:secx = \dfrac{1}{cosx}}}$

$\boxed{ \red{ \sf \: {sin}^{2}x + {cos}^{2}x = 1}}$

$\boxed{ \red{ \sf \: {cos}^{2}x - {sin}^{2}y =cos(x + y)cos(x - y)}}$

$\boxed{ \red{ \sf \:cos90\degree = 0}}$

Now,

Consider,

$\rm :\longmapsto\: \sf \: {sec}^{2}A + {sec}^{2}B = {sec}^{2}A {sec}^{2}B \:$

$\rm :\longmapsto\:\dfrac{1}{ {cos}^{2}A } + \dfrac{1}{ {cos}^{2}B } = \dfrac{1}{ {cos}^{2}A } \times \dfrac{1}{ {cos}^{2}B }$

$\rm :\longmapsto\:\dfrac{ {cos}^{2}B + {cos}^{2}A }{ {cos}^{2}A \: {cos}^{2} B} = \dfrac{1}{ {cos}^{2}A } \times \dfrac{1}{ {cos}^{2}B }$

$\rm :\longmapsto\: {cos}^{2} A + {cos}^{2} B = 1$

$\rm :\longmapsto\: {cos}^{2} A + {cos}^{2} B - 1 = 0$

$\rm :\longmapsto\: {cos}^{2} A - (1 - {cos}^{2} B)= 0$

$\rm :\longmapsto\: {cos}^{2} A - {sin}^{2} B= 0$

$\rm :\longmapsto\:cos(A + B)cos(A - B) = 0$

$\rm :\implies\:cos(A + B) = 0 \: \: \: or \: \: \: \: cos(A - B) = 0$

$\rm :\implies\:A + B = 90\degree \: \: \: or \: \: \: A - B = 90\degree$

$\rm :\implies\:A + B = 90\degree \: \: \: or \: \: \: A =B + 90\degree$

$\rm :\implies\: \triangle \: ABC \: is \: right \: angled \: or \: obtuse \: angled.$

Additional Information :-

Trigonometry Formulas

sin(−θ) = −sin θ

cos(−θ) = cos θ

tan(−θ) = −tan θ

cosec(−θ) = −cosecθ

sec(−θ) = sec θ

cot(−θ) = −cot θ

Product to Sum Formulas

sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

cos x cos y = 1/2[cos(x–y) + cos(x+y)]

sin x cos y = 1/2[sin(x+y) + sin(x−y)]

cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas

sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Sum or Difference of angles

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

sin (A+B) = sin A cos B + cos A sin B

sin (A -B) = sin A cos B – cos A sin B

tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]

tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]

cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A

sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

Multiple and Submultiple angles

sin2A = 2sinA cosA = [2tan A /(1+tan²A)]

cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]

tan 2A = (2 tan A)/(1-tan²A)